Ordinary Differential Equations

$F(x, y, y^{\prime}, y^{\prime \prime}, ..., y^{(n)})$

$y = f(x)$

• $y^{\prime \prime} - 3 y^{\prime} + x y = y^2$

• $y^{\prime} + y = 0$

• $4y^{(5)} - y + x = 0$

$2y - y^2 = 0$

• $y^{\prime} = 3$,

$y = 3x + c$.

• $y^{\prime} - y = 0$,

$y^{\prime} = y$,

$y = ce^x$.

• $y^{\prime} = e^x$,

$y = e^x + c$.

• $y^{\prime} = 4e^x$,

$y = 4e^x$.

$(e^u)^{\prime} = u^{\prime} e^u$.

$(e^x)^{\prime} = 1 e^x = e^x$,

$(kf)^{\prime} = kf^{\prime}$.

Separable Equations

$f(y) \ dy = f(x) \ dx$

$y^{\prime} = \frac{2x}{3y^2} \longrightarrow \frac{dy}{dx} = \frac{2x}{3y^2} \longrightarrow 3y^2 \ dy = 2x \ dx$,

$\int 3y^2 \ dy = \int 2x \ dx \longrightarrow y^3 = x^2 + c \longrightarrow y = \sqrt[3]{x^2 + c}$.

$y^{\prime} = e^{x + y} \longrightarrow \frac{dy}{dx} = e^x \ e^y \longrightarrow \frac{dy}{e^y} = e^x \ dx$,

$e^{-y} \ dy = e^x \ dx \longrightarrow \int e^{-y} \ dy = \int e^x \ dx \longrightarrow -e^{-y} = e^x + c$,

$e^{-y} = c - e^x \longrightarrow ln(e^{-y}) = ln(c - e^x) \longrightarrow -y = ln(c - e^x)$,

$y = -ln(c - e^x)$.

$ln(x) = log_e^x$,

$ln(7) = log_e^7 = 1.94591014906 \longrightarrow e^{1.94591014906} = 7$,

$log_a^b = log_a^b$,

$a^{log_a^b} = b$,

$ln(e^{-y}) = log_e^{e^{-y}}$.

$log_a^{b^n} = n \ log_a^b$,

$log_e^{e^{-y}} = -y \ log_e^e = -y$.

$\int e^{-y}$

$u = -y \longrightarrow du = -dy \longrightarrow -\int e^u \ du = -e^u = -e^{-y}$.

$sec^2(x) \ tan(y) \ dx + sec^2(y) \ tan(x) \ dy = 0$

$sec(x) = \frac{1}{cos(x)}$

$sec^2(x) \ tan(y) \ dx = -sec^2(y) \ tan(x) \ dy$

$\frac{sec^2(x) dx}{tan(x)} = \frac{-sec^2(y) dy}{tan(y)}$

$\int \frac{sec^2(x)}{tan(x)} dx = -\int \frac{sec^2(y)}{tan(y)} dy \longrightarrow ln|tan(x)| = -ln|tan(y)| + c$

$\int \frac{u^{\prime}}{u} du = ln|u| + c$

$(tan(x))^{\prime} = sec^2(x)$

$(tan(u))^{\prime} = u^{\prime} (1 + tan^2(u)) \longrightarrow (tan(x))^{\prime} = 1 + tan^2(x)$

$1 + tan^2(x) = 1 + \frac{sin^2(x)}{cos^2(x)} = \frac{cos^2(x) + sin^2(x)}{cos^2(x)} = \frac{1}{cos^2(x)} = sec^2(x)$

$ln|tan(x)| + ln|tan(y)| = c \longrightarrow ln|tan(y)| = c - ln|tan(x)|$

$ln|tan(x) \ tan(y)| = c \longrightarrow e^{ln|tan(x) \ tan(y)|} = e^c$

$tan(x) \ tan(y) = e^c \longrightarrow tan(y) = \frac{e^c}{tan(x)} \longrightarrow y = tan^{-1} (\frac{e^c}{tan(x)})$.

Exact Equations

$M(x, y) \ dx + N(x, y) \ dy = 0$

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

$f(x, y) = \int M(x, y) \ dx + h(y)$

$N(x, y) = \frac{\partial f(x, y)}{\partial y}$

$(x + y + 1) dx + (x - y^2 + 3) dy = 0$

$\left\{ \begin{array}{l} M(x, y) = x + y + 1 &\\ N(x, y) = x - y^2 + 3 \end{array} \right.$

$\left\{ \begin{array}{l} \frac{\partial M(x, y)}{\partial y} = 1 &\\ \frac{\partial N(x, y)}{\partial x} = 1 \end{array} \right.$

$f(x, y) = \int (x + y + 1) dx + h(y) = \frac{x^2}{2} + y x + x + c + h(y)$

$\frac{\partial f(x, y)}{\partial y} = x + h^{\prime}(y)$

$x + h^{\prime}(y) = x - y^2 + 3 \longrightarrow h^{\prime}(y) = x - y^2 + 3 - x$

$h(y) = \int (-y^2 + 3) dy = \frac{-y^3}{3} + 3y$

$f(x, y) = \frac{x^2}{2} + yx + x - \frac{y^3}{3} + 3y + c$.

$(2x^2 + 2xy^2 + 4y) dx + (2x^2y + 4x + 5y^4) dy = 0$

$\left\{ \begin{array}{l} M(x, y) = 2x^2 + 2xy^2 + 4y &\\ N(x, y) = 2x^2y + 4x + 5y^4 \end{array} \right.$

$\left\{ \begin{array}{l} \frac{\partial M(x, y)}{\partial y} = 4xy + 4 &\\ \frac{\partial N(x, y)}{\partial x} = 4xy + 4 \end{array} \right.$

$f(x, y) = \int (2x^2 + 2xy^2 + 4y) dx + h(y)$

$f(x, y) = \frac{2}{3} x^3 + x^2y^2 + 4yx + c + h(y)$

$N(x, y) = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (\frac{2}{3} x^3 + x^2y^2 + 4yx + c + h(y))$

$N(x, y) = 2x^2y + 4x + h^{\prime}(y) = 2x^2 y + 4x + 5y^4$

$h^{\prime}(y) = 2x^2 y + 4x + 5y^4 - 4x - 2x^2y = 5y^4$

$h(y) = \int h^{\prime}(y) dy = \int 5y^4 dy = y^5 + c$

$f(x, y) = \frac{2}{3} x^3 + x^2y^2 + 4yx + y^5 + c$.

$f(tx, ty) = t^n f(x, y)$,

$n \in \mathbb{Z}$,

$t \neq 0$.

$f(x, y) = x^4 - x^3 y$

$f(tx, ty) = (tx)^4 - (tx)^3 (ty) = t^4 x^4 - t^3 x^3 t y = t^4 x^4 - t^4 x^3 y = t^4 (x^4 - x^3 y) = t^4 f(x, y)$

$n = 4$

First-Order Homogeneous Equations

$M(x, y) dx + N(x, y) dy = 0$

$u = \frac{y}{x}$

$y^{\prime} = 1 + \frac{y}{x}$

$y^{\prime} = \frac{x + y}{x}$

$M(tx, ty) = tx + ty = t(x + y) = t^1 M(x, y)$

$\left\{ \begin{array}{l} M(x, y) dx = N(x, y) dy &\\ (x + y) dx = (x) dy \end{array} \right.$

$y = ux \longrightarrow dy = x \ du + u \ dx$

$(x + ux) dx - x(x \ du + u \ dx) = 0$

$x \ dx + u x \ dx - x^2 \ du - x u \ dx = 0$

$x \ dx = x^2 \ du \longrightarrow \frac{1}{x} dx = du$

$\int \frac{1}{x} dx = \int du$,

$ln(x) = u + c \longrightarrow ln(x) = \frac{y}{x} + c \longrightarrow ln(x) = \frac{y + cx}{x}$,

$y = x \ ln(x) - cx$.

$(x e^{\frac{y}{x}} + y) dx - x \ dy = 0$.

$y = ux \longrightarrow dy = x \ du + u \ dx \longrightarrow (x e^u + ux) dx - x (x \ du + u \ dx) = 0$

$x e^u \ dx + ux \ dx - x^2 \ du - xu \ dx = 0$

$x e^u \ dx = x^2 \ du \longrightarrow \frac{1}{x} dx = e^{-u} du$

$\int \frac{1}{x} dx = \int e^{-u} du \longrightarrow ln(x) = -e^{-u} + c$

$ln(x) = -e^{-\frac{y}{x}} + c$

$ln(c - ln(x)) = ln(e^{-\frac{y}{x}}) \longrightarrow ln(c - ln(x)) = -ln(e^{\frac{y}{x}}) \longrightarrow -ln(c - ln(x)) = \frac{y}{x}$

$y = -x \ ln(c - ln(x))$.

$log_b^a = log_b^a$,

$b^{log_b^a} = a$,

$log_e^{e^y} = y \ log_e^e = y$.

First-Order Linear Equations

$y^{\prime} + p(x) y = q(x)$

$y = \frac{1}{e^{\int p(x)dx}}(\int e^{\int p(x)dx} q(x)dx + c)$

$y^{\prime} + y = e^x$

$\left\{ \begin{array}{l} p(x) = 1 &\\ q(x) = e^x \end{array} \right.$

$\int p(x) dx = \int dx = x + c$

$y = \frac{1}{e^{x + c}} (\int e^{x + c} e^x dx + c)$

$y = \frac{1}{e^{x + c}} (\int e^{2x + c} dx + c) = \frac{1}{e^{x + c}} (\frac{e^{2x + c}}{2} + c)$

$y = \frac{e^x}{2} + ce^{-x - c}$

$y = \frac{1}{e^x} (\frac{1}{2} e^{2x} + c)$.

$y^{\prime} + \frac{2x}{1 + x^2} y = \frac{cot(x)}{1 + x^2}$

$\left\{ \begin{array}{l} p(x) = \frac{2x}{1 + x^2} &\\ q(x) = \frac{cot(x)}{1 + x^2} \end{array} \right.$

$y^{\prime} + p(x) y = q(x)$

$e^{\int p(x) dx} = e^{\int \frac{2x}{1 + x^2} dx} = e^{\int \frac{du}{u}} = e^{ln|u|} = u$.

$e^{\int p(x) dx} = u = 1 + x^2$

$y = \frac{1}{1 + x^2} (\int q(x) e^{\int p(x) dx} dx + c)$

$y = \frac{1}{1 + x^2} (\int \frac{cot(x)}{1 + x^2} (1 + x^2) dx + c)$

$y = \frac{1}{1 + x^2} (\int cot(x) dx + c) = \frac{1}{1 + x^2} (ln(|sin(x)|) + c)$

$\int cot(x) dx = \int \frac{cos(x)}{sin(x)} dx = ln|sin(x)| + c$.

Bernoulli Equations

$y^{\prime} + p(x) y = q(x) y^n$.

$n = 0, n = 1$

$n \neq 0, n \neq 1$

$/ y^n$

$u = \frac{1}{y^{n - 1}}$

$y^{\prime} + \frac{1}{x} y = y^2$

$n = 2 \longrightarrow y^n = y^2$

$\frac{y^{\prime}}{y^2} + \frac{1}{xy} = 1$

$\left\{ \begin{array}{l} u = \frac{1}{y} = y^{-1} &\\ du = d(y^{-1}) = -y^{-2} dy \end{array} \right.$

$\frac{y^{\prime}}{y^2} + \frac{1}{x} \frac{1}{y} = 1$

$y^{\prime} u^2 + \frac{u}{x} = 1$

$\left\{ \begin{array}{l} -du = dy \ y^{-2} &\\ u^{\prime} = - y^{\prime} y^{-2} \end{array} \right.$

$-u^{\prime} + \frac{u}{x} = 1$

$u^{\prime} - \frac{u}{x} = -1 \longrightarrow u^{\prime} - \frac{u}{x} + 1 = 0$

$\left\{ \begin{array}{l} p(x) = \frac{-1}{x} &\\ q(x) = -1 \end{array} \right.$

$e^{\int p(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-ln|x|} = \frac{1}{x}$

$u = \frac{1}{e^{\int p(x) dx}} (\int q(x) e^{\int p(x) dx} dx + c)$

$u = x (\int (-1) \frac{1}{x} dx + c) = x(\int -\frac{1}{x} dx + c)$

$u = x (-ln|x| + c)$

$u = \frac{1}{y}$

$y^{-1} = u = x(-ln|x| + c) \longrightarrow y = \frac{1}{x(-ln|x| + c)}$.

$(y^n)^{\prime} = n y^{\prime} y^{n - 1}$

$y = x \longrightarrow (x^n)^{\prime} = n x^{\prime} x^{n - 1}$.

Review Exercises

• $y^\prime = \frac{5x}{2y^2}$

$\frac{dy}{dx} = \frac{5x}{2y^2} \longrightarrow 5x \ dx = 2y^2 \ dy$

$\int 5x \ dx = \int 2y^2 \ dy \longrightarrow \frac{5x^2}{2} = \frac{2y^3}{3} + c$

$\frac{2y^3}{3} = \frac{5x^2}{2} + c \longrightarrow y^3 = \frac{3}{2} (\frac{5x^2}{2} - c)$

$y = \sqrt[3]{\frac{3}{2} (\frac{5x^2}{2} - c)}$.

• $xy^\prime + y = y^2$

$x \frac{dy}{dx} + y = y^2 \longrightarrow \frac{x}{dx} + \frac{y}{dy} = \frac{y^2}{dy}$

$\frac{x}{dx} = \frac{y^2}{dy} - \frac{y}{dy} \longrightarrow \frac{x}{dx} = \frac{y^2 - y}{dy}$

$\frac{dx}{x} = \frac{dy}{y^2 - y} \longrightarrow \frac{dx}{x} = \frac{dy}{y (y - 1)}$

$\frac{dx}{x} = dy (\frac{A}{y} + \frac{B}{y - 1}) \longrightarrow \frac{dx}{x} = \frac{-dy}{y} + \frac{dy}{y - 1}$

$\frac{A (y - 1) + B y}{y (y - 1)} = \frac{1}{y (y - 1)}$

$Ay - A + By = 1$

$\left\{ \begin{array}{l} A + B = 0 &\\ -A = 1 \end{array} \right.$

$\left\{ \begin{array}{l} A = -1 &\\ B = 1 \end{array} \right.$

$\int \frac{dx}{x} = - \int \frac{dy}{y} + \int \frac{dy}{y - 1} \longrightarrow ln|x| = -ln|y| + ln|y - 1| + c$

$ln|x| = ln|\frac{y - 1}{y}| + c \longrightarrow e^{ln|x|} = e^{ln|\frac{y - 1}{y}| + c} \longrightarrow x = \frac{y - 1}{y} + c_1$

$x = 1 - \frac{1}{y} + c_1 \longrightarrow \frac{1}{y} = 1 - x + c_1$

$y = \frac{1}{1 - x + c_1}$.

• $(x^2 + y^2) dx + 2xy \ dy = 0$

$u = \frac{y}{x} \longrightarrow y = ux \longrightarrow dy = u \ dx + x \ du$

$(x^2 + u^2 x^2) dx + 2x ux \ dy = 0$

$x^2 (u^2 + 1) dx + 2x^2 u \ dy = 0 \longrightarrow x^2 (u^2 + 1) dx + 2x^2 u (u \ dx + x \ du) = 0$

$(u^2 + 1) dx + 2u^2 \ dx + 2ux \ du = 0$

$dx(u^2 + 1 + 2u^2) + 2ux \ du = 0 \longrightarrow (3u^2 + 1) dx + 2ux \ du = 0$

$\frac{3u^2 + 1}{x} dx + 2u \ du = 0 \longrightarrow \frac{dx}{x} + \frac{2u}{3u^2 + 1} du = 0$

$\frac{dx}{x} = \frac{-2u}{3u^2 + 1} du \longrightarrow \int \frac{dx}{x} = \int \frac{-2u}{3u^2 + 1} du$

$ln|x| = \frac{-1}{3} \int \frac{2u}{u^2 + 1} du \longrightarrow ln|x| = \frac{-1}{3} ln|u^2 +1| + c$

$e^{ln|x|} = e^{\frac{-1}{3} ln|u^2 + 1| + c}$

$x = e^{ln|(u^2 + 1)^{\frac{-1}{3}}| + c} \longrightarrow x = \frac{1}{\sqrt[3]{u^2 + 1}} + c_1$

$\frac{1}{x} = \sqrt[3]{u^2 + 1} + c_1 \longrightarrow \frac{1}{x^3} = u^2 + 1 + c_1$

$u^2 = \frac{1}{x^3} - 1 - c_1 \longrightarrow u = \sqrt{\frac{1}{x^3} - 1 - c_1}$

$\frac{y}{x} = \sqrt{\frac{1}{x^3} - 1 - c_1}$

$y = x \sqrt{\frac{1}{x^3} - 1 - c_1}$.

• $y^\prime = 2 + \frac{y}{x}$

$u = \frac{y}{x} \longrightarrow y = ux \longrightarrow dy = u \ dx + x \ du$

$\frac{dy}{dx} = u + x \frac{du}{dx} \longrightarrow y^\prime = u + x u^\prime$

$u + xu^\prime = 2 + u \longrightarrow x u^\prime = 2 \longrightarrow u^\prime = \frac{2}{x}$

$\frac{du}{dx} = \frac{2}{x} \longrightarrow du = \frac{2 dx}{x}$

$\int du = \int 2 \frac{dx}{x} \longrightarrow u = 2 ln|x| + c$

$\frac{y}{x} = ln(x^2) + c \longrightarrow y = x \ ln(x^2) + cx$.

• $(x + y + 2) dx + (x - y^2 + 1) dy = 0$

$f(x, y) = \int (x + y + 2) dx + h(y) = \frac{x^2}{2} + xy + 2x + c + h(y)$

$\frac{\partial f}{\partial y} = N(x, y)$

$x + h^\prime (y) = x - y^2 + 1 \longrightarrow h^\prime (y) = \int h(y) dy = \int (1 - y^2) dy = y - \frac{y^3}{3} + c_1$

$f(x, y) = \frac{x^2}{2} + xy + 2x + y - \frac{y^3}{3} + c_2$.

• $(2x^2 + 4y) dx + (4x - 3y^2) dy = 0$

$f(x, y) = \int M(x, y) dx + h(y) \longrightarrow f(x, y) = \int (2x^2 + 4y) dx + h(y) = \frac{2x^3}{3} + 4xy + h(y)$

$\frac{\partial f(x, y)}{\partial y} = N(x, y) \longrightarrow 4x + h^\prime (y) = 4x - 3y^2 \longrightarrow h^\prime (y) = -3y^2$

$h(y) = \int h^\prime (y) dy \longrightarrow h(y) = \int -3y^2 \ dy \longrightarrow h(y) = -3 \frac{y^3}{3} + c \longrightarrow h(y) = -y^3 + c$

$f(x, y) = \frac{2}{3} x^3 + 4xy - y^3 + c$.

• $(2xe^y + e^x) dx + (x^2 + 1) e^y dy = 0$

$f(x, y) = \int M(x, y) dx + h(y) = \int (2xe^y + e^x) dx + h(y) = \frac{2x^2}{2} e^y + e^x + c + h(y)$

$\frac{\partial f(x, y)}{\partial y} = N(x, y) \longrightarrow x^2 e^y + h^\prime (y) = e^y (x^2 + 1)$

$h^\prime (y) = e^y (x^2 + 1) - x^2 e^y = e^y x^2 + e^y - e^y x^2 = e^y$

$h(y) = \int h^\prime (y) dy = \int e^y dy = e^y + c_1$

$f(x, y) = x^2 e^y + e^x + e^y + c_2$

• $\frac{dy}{dx} + y = e^x$

$\left\{ \begin{array}{l} p(x) = 1 &\\ q(x) = e^x \end{array} \right.$

$\int p(x) dx = \int dx = x$

$y = \frac{1}{e^{\int p(x) dx}} (\int e^{\int p(x) dx} q(x) dx + c)$

$y = \frac{1}{e^x} (\int e^x e^x dx + c) \longrightarrow y = e^{-x} (\int e^{2x} dx + c)$

$y = e^{-x} (\frac{e^{2x}}{2} + c) \longrightarrow y = \frac{e^x}{2} + c e^{-x}$.

• $\frac{dy}{dx} = \frac{e^{2y}}{xe^{2y} - y}$

$\frac{dx}{dy} = \frac{x e^{2y} - y}{e^{2y}} = \frac{x e^{2y}}{e^{2y}} - \frac{y}{e^{2y}}$

$\frac{dx}{dy} = x - \frac{y}{e^{2y}} \longrightarrow x^\prime - x = -\frac{y}{e^{2y}}$

$\left\{ \begin{array}{l} p(y) = -1 &\\ q(y) = \frac{-y}{e^{2y}} \end{array} \right.$

$\int p(y) dy = \int -dy = -y$

$x = \frac{1}{e^{\int p(y) dy}} (\int e^{\int p(y) dy} q(y) dy + c)$

$x = \frac{1}{e^{-y}} (\int e^{-y} \frac{-y}{e^{2y}} dy + c) \longrightarrow x = e^y (\int -e^{-3y} y \ dy + c)$

$\left\{ \begin{array}{l} u = y \longrightarrow du = dy &\\ dv = -e^{-3y} dy \longrightarrow v = \frac{e^{-3y}}{3} \end{array} \right.$

$\int -e^{-3y} y \ dy = \int u \ dv = uv - \int v \ du = y \frac{e^{-3y}}{3} - \int \frac{e^{-3y}}{3} dy$

$\int -e^{-3y} y \ dy = \frac{y}{3e^{3y}} - (\frac{-1}{3})) \frac{e^{-3y}}{3} = \frac{y}{3e^{3y}} + \frac{1}{9} e^{-3y}$

$x = e^y (\frac{y}{3} e^{-3y} + \frac{1}{9} e^{-3y} + c) \longrightarrow x = e^{-2y} (\frac{y}{3} + \frac{1}{9}) + c e^y$.

• $y^\prime - \frac{y}{x} = y^2$

$n = 2$

$y^\prime y^{-2} - \frac{1}{xy} = 1$

$u = \frac{1}{y^{2 - 1}} = \frac{1}{y} \longrightarrow y = \frac{1}{u} \longrightarrow y^\prime = (u^{-1})^\prime = -u^{-2} u^\prime$

$u^\prime (-u^{-2}) (u^2) - \frac{u}{x} = 1$

$-u^\prime - \frac{u}{x} = 1 \longrightarrow u^\prime + \frac{u}{x} = -1$

$\left\{ \begin{array}{l} p(x) = \frac{1}{x} &\\ q(x) = -1 \end{array} \right.$

$\int p(x) dx = \int \frac{dx}{x} = ln|x|$

$u = \frac{1}{e^{\int p(x) dx}} (\int e^{\int p(x) dx} q(x) dx + c)$

$u = \frac{1}{e^{ln|x|}} (\int e^{ln|x|} (-1) dx + c)$

$u = \frac{1}{x} (\int -\frac{dx}{x} + c) = \frac{1}{x} (-ln|x| + c)$

$\frac{1}{y} = \frac{-ln|x| + c}{x}$

$y = \frac{x}{-ln|x| + c}$.

• $y^\prime + xy = \frac{x}{y^3}$

$n = 3$

$u = \frac{1}{y^{3 - 1}} = \frac{1}{y^2} \longrightarrow y^{-2} = u \longrightarrow -2y^{-3}dy = du \longrightarrow dy = \frac{du}{-2}y^3$

$y^\prime y^{-3} + xy^{-2} = x$

$\frac{-u^\prime}{2} + xu = x \longrightarrow u^\prime - 2xu = -2x$

$\left\{ \begin{array}{l} p(x) = -2x &\\ q(x) = -2x \end{array} \right.$

$\int p(x) dx = \int -2x \ dx = -x^2$

$u = \frac{1}{e^{\int p(x) dx}} (\int e^{\int p(x) dx} q(x) dx + c)$

$u = \frac{1}{e^{-x^2}} (\int e^{-x^2} (-2x) dx + c)$

$\left\{ \begin{array}{l} v = -x^2 &\\ dv = -2x \ dx \end{array} \right.$

$\int -e^{-x^2} 2x \ dx = \int e^v dv = e^v = e^{-x^2}$

$u = \frac{1}{e^{-x^2}} (e^{-x^2} + c) = 1 + c e^{x^2}$

$\frac{1}{y^2} = 1 + c e^{x^2} \longrightarrow y^2 = \frac{1}{1 + c e^{x^2}}$

$y = \sqrt{\frac{1}{1 + c e^{x^2}}}$.

Second-Order Linear Homogeneous Equations (with Constant Coefficients)

$y^{\prime \prime} + a y^{\prime} + by = 0$

$r^2 + ar + b = 0$

$\Delta > 0 \longrightarrow y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$.

$y^{\prime \prime} + y^{\prime} - 2y = 0$

$\left\{ \begin{array}{l} a = 1 &\\ b = -2 \end{array} \right.$

$r^2 + r - 2 = 0$,

$\Delta = 1^2 - 4 (1) (-2) = 9 > 0$,

$r_1 = \frac{-1 + \sqrt{9}}{2 (1)} = 1$,

$r_2 = \frac{-1 - \sqrt{9}}{2 (1)} = -2$,

$y = c_1 e^{x} + c_2 e^{-2x}$.

$\Delta = 0 \longrightarrow y = (c_1 + c_2 x) e^{r x}$.

$y^{\prime \prime} + 4 y^{\prime} + 4y = 0$.

$r^2 + 4r + 4 = 0$

$\Delta = 4^2 - 4 (1) (4) = 0 \longrightarrow r = \frac{-4}{2 (1)} = -2$

$y = (c_1 + c_2 x) e^{-2x}$.

$\Delta < 0$

$i^2 = -1 \longrightarrow i = \sqrt{-1}$

$z^2 = -1 \longrightarrow z^2 + 1 = 0$

$i \in \mathbb{C}$

$\left\{ \begin{array}{l} r_1 = \alpha + \beta i &\\ r_2 = \alpha - \beta i \end{array} \right.$

$r_1, r_2 \in \mathbb{C} = \{ x + y i | x \in \mathbb{R}, y \in \mathbb{R} \}$

$y = e^{\alpha x} (c_1 cos(\beta x) + c_2 sin(\beta x))$

$y^{\prime \prime} + 4 y^{\prime} + 5y = 0$

$r^2 + 4r + 5 = 0$,

$\Delta = 16 - 20 = -4$,

$r_1 = \frac{-4 + \sqrt{-4}}{2} = -2 + i$,

$r_2 = \frac{-4 - \sqrt{-4}}{2} = -2 - i$,

$\left\{ \begin{array}{l} \alpha = -2 &\\ \beta = 1 \end{array} \right.$,

$y = e^{-2x} (c_1 cos(x) + c_2 sin(x))$.

Inexact Equations

$\left\{ \begin{array}{l} M(x, y) dx + N(x, y) dy = 0 &\\ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \end{array} \right.$

$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$

$h(x, y)$

$h(x, y) M(x, y) dx + h(x, y) N(x, y) dy = 0$

$\left\{ \begin{array}{l} M(x, y) dx + N(x, y) dy = 0 &\\ \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \end{array} \right.$

$p(x) = \frac{1}{N} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$

$e^{\int p(x) dx}$.

$p(y) = \frac{-1}{M} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$

$e^{\int p(y) dy}$.

$dx + \frac{x - sin(y)}{y} dy = 0$.

$\left\{ \begin{array}{l} M(x, y) = 1 &\\ N(x, y) = \frac{x - sin(y)}{y} \end{array} \right.$

$\left\{ \begin{array}{l} \frac{\partial M}{\partial y} = 0 &\\ \frac{\partial N}{\partial x} = \frac{1}{y} \end{array} \right.$

$\left\{ \begin{array}{l} 0 \neq \frac{1}{y} &\\ \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \end{array} \right.$

$p(y) = \frac{-1}{1} (0 - \frac{1}{y}) = \frac{1}{y}$

$e^{\int p(y) dy} = e^{\int \frac{1}{y} dy} = e^{ln|y|} = y$

$y \ dx + (x - sin(y)) dy = 0$

$\left\{ \begin{array}{l} M_1(x, y) = y &\\ N_1(x, y) = x - sin(y) \end{array} \right.$

$\left\{ \begin{array}{l} \frac{\partial M_1}{\partial y} = 1 &\\ \frac{\partial N_1}{\partial x} = 1 \end{array} \right.$

$\frac{\partial M_1}{\partial y} = \frac{\partial N_1}{\partial x}$

$f(x, y) = \int M_1(x, y) dx + h(y) = \int y \ dx + h(y)$

$f(x, y) = xy + c + h(y)$

$\frac{\partial f(x, y)}{\partial y} = N_1(x, y) = \frac{\partial}{\partial y} (xy + c + h(y))$

$N_1(x, y) = x + h^{\prime}(y)$,

$x + h^{\prime}(y) = x - sin(y) \longrightarrow h^{\prime}(y) = -sin(y)$,

$f(x, y) = xy + c + \int h^{\prime}(y) dy = xy + c + \int -sin(y) dy$,

$f(x, y) = xy + c + cos(y) + c_1$,

$f(x, y) = xy + cos(y) + c_2$.

$(x^2 + x - y^2) dx + x y \ dy = 0$.

$\left\{ \begin{array}{l} M(x, y) = x^2 + x - y^2 &\\ N(x, y) = xy \end{array} \right.$

$\left\{ \begin{array}{l} \frac{\partial M(x, y)}{\partial y} = -2y &\\ \frac{\partial N(x, y)}{\partial x} = y \end{array} \right.$

$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$

$\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = -2y - y = -3y$

$p(x) = \frac{1}{N} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = \frac{1}{xy} (-3y) = \frac{-3}{x}$

$e^{\int p(x) dx} = e^{-3 \int \frac{dx}{x}} = e^{-3 ln|x|} = e^{ln|x^{-3}|} = x^{-3}$

$x^{-3} (x^2 + x - y^2) dx + x^{-3} (xy) dy = 0$

$(x^{-1} + x^{-2} - y^2 x^{-3}) dx + (x^{-2} y) dy = 0$

$\left\{ \begin{array}{l} M_1(x, y) = x^{-1} + x^{-2} - y^2 x^{-3} &\\ N_1(x, y) = x^{-2} y \end{array} \right.$

$\frac{\partial M_1}{\partial y} = -2y x^{-3}$

$\frac{\partial N_1}{\partial x} = -2x^{-3} y$

$\frac{\partial M_1}{\partial y} = \frac{\partial N_1}{\partial x}$

$f(x, y) = \int M_1(x, y) dx + h(y) = \int x^{-1} + x^{-2} - y^2 x^{-3} \ dx + h(y)$

$f(x, y) = ln|x| - \frac{1}{x} + y^2 \frac{x^{-2}}{2} + c + h(y)$

$N_1(x, y) = \frac{\partial f(x, y)}{\partial y} = \frac{\partial}{\partial y} (ln|x| - \frac{1}{x} + y^2 \frac{x^{-2}}{2} + c + h(y))$

$N_1(x, y) = yx^{-2} + h^{\prime}(y)$,

$h^{\prime}(y) = 0 \longrightarrow h(y) = \int h^{\prime}(y) dy = \int 0 dy = c_1$,

$f(x, y) = ln|x| - \frac{1}{x} + y^2 \frac{x^{-2}}{2} + c + c_1$,

$f(x, y) = ln|x| - \frac{1}{x} + y^2 \frac{x^{-2}}{2} + c_2$.

Laplace Transform

$D: f(x) \to f^{\prime}(x)$,

$D f(x) = f^{\prime}(x)$.

$L\{ f(t) \} = \int_0^\infty e^{-st} f(t) dt = F(s)$.

$f(t) = 1$

$L\{ f(t) \} = \int_0^{\infty} e^{-st} dt = -\frac{1}{s} e^{-st} |_0^\infty = 0 - (-\frac{1}{s}) = \frac{1}{s}$.

$s > 0$

$f(t) = e^{at}$

$L\{ e^{at} \} = \int_0^\infty e^{-st} e^{at} dt = \int_0^\infty e^{(a - s) t} dt$

$L\{ e^{at} \} = \frac{e^{(a - s) t}}{a - s} |_0^\infty = \frac{e^{(a - s) \infty}}{a - s} - \frac{e^0}{a - s}$

$\left\{ \begin{array}{l} a > s \longrightarrow \infty &\\ a < s \longrightarrow -\frac{1}{a - s} = \frac{1}{s - a} \end{array} \right.$

$\int e^{at} dt$

$u = at \longrightarrow du = a \ dt$

$\frac{1}{a} \int e^u du = \int e^{at} dt = \frac{1}{a} e^u + c = \frac{1}{a} e^{at} + c$

$f(t) \to L \{ f(t) \} = F(s)$:

$f(t) = 1 \to L\{ 1 \} = F(s) = \frac{1}{s}, \ D_F: s > 0$,

$f(t) = e^{at} \to L\{ e^{at} \} = F(s) = \frac{1}{s - a}, \ D_F: s > 0$,

$f(t) = t \to L\{ t \} = F(s) = \frac{1}{s^2}, \ D_F: s > 0$,

$f(t) = t^n \ (n \in \mathbb{N}) \to L\{ t^n \} = F(s) = \frac{n!}{s^{n + 1}}, \ D_F: s > 0$,

$f(t) = cos(at) \to L\{ cos(at) \} = F(s) = \frac{s}{s^2 + a^2}, \ D_F: s > 0$,

$f(t) = sin(at) \to L\{ sin(at) \} = F(s) = \frac{a}{s^2 + a^2}, \ D_F: s > 0$.

$F(s) \leftrightarrow f(t)$

$F(s) = \frac{1}{s - 8}$

$L^{-1}\{ \frac{1}{s - 8} \} = e^{8t} = f(t)$

$F(s) = \frac{5s - 1}{s^2 - 1}$

$\frac{5s - 1}{s^2 - 1} = \frac{5s - 1}{(s + 1) (s - 1)} = \frac{A}{s - 1} + \frac{B}{s + 1} = \frac{A (s + 1) + B (s - 1)}{(s - 1) (s + 1)}$

$\frac{As + Bs + A - B}{(s - 1) (s + 1)} = \frac{(A + B) s + (A - B)}{(s - 1) (s + 1)}$

$\left\{ \begin{array}{l} A + B = 5 &\\ A - B = -1 \end{array} \right.$

$\left\{ \begin{array}{l} 2A = 4 \longrightarrow A = 2 &\\ B = 3 \end{array} \right.$

$L^{-1} \{ \frac{5s - 1}{s^2 - 1} \} = L^{-1} \{ \frac{2}{s - 1} + \frac{3}{s + 1} \} = 2 e^t + 3 e^{-t}$.

$L^{-1} \{ \frac{3s - 2}{s^3} \}$

$L^{-1} \{ \frac{3s - 2}{s^3} \} = L^{-1} \{ \frac{A}{s^3} + \frac{B}{s^2} + \frac{C}{s} \} = L^{-1} \{ \frac{A + Bs + Cs^2}{s^3} \}$

$3s - 2 = A + Bs + C s^2$

$\left\{ \begin{array}{l} A = -2 &\\ B = 3 &\\ C = 0 \end{array} \right.$

$L^{-1} \{ \frac{3s - 2}{s^3} \} = L^{-1} \{ \frac{-2}{s^3} + \frac{3}{s^2} + \frac{0}{s} \} = L^{-1} \{ \frac{-2}{s^3} \} + L^{-1} \{ \frac{3}{s^2} \} = -t^2 + 3t$

$t^n \leftrightarrow \frac{n!}{s^{n + 1}}$

$L\{ f(t) \}, \ L\{ f^{\prime}(t) \}$

$L \{ f^{\prime}(t) \} = s L \{ f(t) \} - f(0)$

$L \{ f^{\prime \prime}(t) \} = s L \{ f^\prime (t) \} - f^\prime (0)$

$L \{ f^{\prime \prime} (t) \} = s (s L \{ f(t) \} - f(0)) - f^\prime (0)$

$L \{ f^{\prime \prime} (t) \} = s^2 L \{ f(t) \} - s \ f(0) - f^\prime (0)$

$y^\prime + y = e^{2t}$,

$f(0) = y(0) = 0$.

$L \{ y^\prime + y \} = L \{ e^{2t} \}$

$L \{ y^\prime \} + L \{ y \} = \frac{1}{s - 2}$

$s \ L \{ y \} - y(0) + L \{ y \} = \frac{1}{s - 2}$

$L \{ y \} (s + 1) = \frac{1}{s - 2}$

$L \{ y \} = \frac{1}{(s + 1) (s - 2)} \longrightarrow y = L^{-1} \{ \frac{1}{(s + 1) (s - 2)} \}$

$\frac{1}{(s + 1) (s - 2)} = \frac{A}{s + 1} + \frac{B}{s - 2} = \frac{A (s - 2) + B (s + 1)}{(s + 1) (s - 2)} = \frac{(A + B) s - 2A + B}{(s + 1) (s - 2)}$

$\left\{ \begin{array}{l} A + B = 0 &\\ -2A + B = 1 \end{array} \right.$

$3A = -1 \longrightarrow A = \frac{-1}{3} \longrightarrow B = \frac{1}{3}$

$y = L^{-1} \{ \frac{\frac{-1}{3}}{s + 1} + \frac{\frac{1}{3}}{s - 2} \} = \frac{-1}{3} L^{-1} \{ \frac{1}{s + 1} \} + \frac{1}{3} L^{-1} \{ \frac{1}{s - 2} \}$

$y = \frac{-1}{3} e^{-t} + \frac{1}{3} e^{2t}$.

$L \{ y^\prime \} = s L \{ y \} - y(0)$.

$L \{ y^{\prime \prime} \} = s L \{ y^{\prime} \} - y^\prime (0) = s (s L \{ y \} - y(0)) - y^\prime (0)$,

$L \{ y^{\prime \prime} \} = s^2 L \{ y \} - s \ y(0) - y^{\prime} (0)$.

$y^{\prime \prime} - 4 y^\prime - 5 y = 0$,

$\left\{ \begin{array}{l} y(0) = 1 &\\ y^\prime (0) = 0 \end{array} \right.$.

$L \{ y^{\prime \prime} - 4 y^\prime - 5y \} = L \{ 0 \}$

$L \{ y^{\prime \prime} \} - 4 L \{ y^\prime \} - 5 L \{ y \} = 0$

$s^2 L \{ y \} - s (1) - 4 (s L \{ y \} - 1) - 5 L \{ y \} = 0$

$s^2 L \{ y \} - s - 4s L \{ y \} + 4 - 5 L \{ y \} = 0$

$L \{ y \} (s^2 - 4s - 5) - s + 4 = 0$

$L \{ y \} (s^2 - 4s - 5) = s - 4$

$L \{ y \} = \frac{s - 4}{s^2 - 4s - 5} \longrightarrow y = L^{-1} \{ \frac{s - 4}{s^2 - 4s - 5} \}$

$\frac{s - 4}{s^2 - 4s - 5} = \frac{s - 4}{(s + 1) (s - 5)} = \frac{A}{s + 1} + \frac{B}{s - 5}$

$\frac{A (s - 5) + B (s + 1)}{(s + 1) (s - 5)} = \frac{(A + B) s - 5A + B}{(s + 1) (s - 5)}$

$\left\{ \begin{array}{l} A + B = 1 &\\ -5A + B = -4 \end{array} \right.$

$\left\{ \begin{array}{l} A + B = 1 &\\ 5A - B = 4 \end{array} \right.$

$\left\{ \begin{array}{l} 6A = 5 \longrightarrow A = \frac{5}{6} &\\ B = 1 - \frac{5}{6} = \frac{1}{6} \end{array} \right.$

$y = L^{-1} \{ \frac{\frac{5}{6}}{s + 1} + \frac{\frac{1}{6}}{s - 5} \} = \frac{5}{6} L^{-1} \{ \frac{1}{s + 1} \} + \frac{1}{6} L^{-1} \{ \frac{1}{s - 5} \}$

$y = \frac{5}{6} e^{-t} + \frac{1}{6} e^{5t}$

$y^{\prime \prime} - 4 y^\prime - 5y = 0$

$r^2 - 4r - 5 = 0$

$\Delta = \frac{4 \pm \sqrt{16 + 20}}{2}$

$\Delta = \frac{4 \pm 6}{2} = 2 \pm 3$

$\left\{ \begin{array}{l} r_1 = 5 &\\ r_2 = -1 \end{array} \right.$

$(r + 1) (r - 5) = 0$

$y = c_1 e^{r_1 x} + c_2 e^{r_2 x} \longrightarrow y = c_1 e^{-t} + c_2 e^{5t}$

$\left\{ \begin{array}{l} y(0) = 1 &\\ y^\prime (0) = 0 \end{array} \right.$

$y(0) = 1 \longrightarrow c_1 e^{-(0)} + c_2 e^{5 (0)} = 1 \longrightarrow c_1 + c_2 = 1$

$y^\prime (0) = 0 \longrightarrow y^\prime = -c_1 e^{-x} + 5 c_2 e^{5x} = 0$

$y^\prime (0) = -c_1 e^0 + 5c_2 e^0 = 0 \longrightarrow -c_1 + 5c_2 = 0$

$\left\{ \begin{array}{l} c_1 + c_2 = 1 &\\ -c_1 + 5c_2 = 0 \end{array} \right.$

$6c_2 = 1 \longrightarrow c_2 = \frac{1}{6}$

$c_1 + \frac{1}{6} = 1 \longrightarrow c_1 = 1 - \frac{1}{6} = \frac{5}{6}$

$y = \frac{5}{6} e^{-t} + \frac{1}{6} e^{5t}$.

$y^{\prime \prime} - 2 y^\prime - 8y = 0$,

$\left\{ \begin{array}{l} y(0) = 1 &\\ y^\prime (0) = 0 \end{array} \right.$

$L \{ y^\prime \} = s L \{ y \} - y(0)$

$L \{ y^{\prime \prime} \} = s L \{ y^\prime \} - y^\prime (0) = s (s L \{ y \} - y(0)) - y^\prime (0)$

$L \{ y^{\prime \prime} \} = s^2 L \{ y \} - s \ y(0) - y^\prime (0)$

$s^2 L \{ y \} - s \ y(0) - y^\prime (0) - 2s \ L \{ y \} + 2y(0) - 8 L \{ y \} = 0$.

$L \{ y \} (s^2 - 2s - 8) - s y(0) - y^\prime (0) + 2 y(0) = 0$

$L \{ y \} (s^2 - 2s - 8) - s + 2 = 0 \longrightarrow L \{ y \} (s^2 - 2s - 8) = s - 2$

$L \{ y \} = \frac{s - 2}{s^2 - 2s - 8} \longrightarrow L \{ y \} = \frac{s - 2}{(s - 4) (s + 2)}$

$\sqrt{\Delta} = \sqrt{(-2)^2 - 4 (1) (-8)} = \sqrt{4 + 32} = \sqrt{36} = 6$

$s_{1, 2} = \frac{2 \pm 6}{2} = 1 \pm 3$

$\left\{ \begin{array}{l} s_1 = 4 &\\ s_2 = -2 \end{array} \right.$

$y = L^{-1} \{ \frac{s - 2}{(s - 4) (s + 2)} \} = L^{-1} \{ \frac{A}{s - 4} + \frac{B}{s + 2} \}$

$y = L^{-1} \{ \frac{A (s + 2)}{(s - 4) (s + 2)} + \frac{B (s - 4)}{(s - 4) (s + 2)} \}$

$y = L^{-1} \{ \frac{As + 2A + Bs - 4B}{(s - 4) (s + 2)} \} = L^{-1} \{ \frac{s - 2}{(s - 4) (s + 2)} \}$

$\left\{ \begin{array}{l} A + B = 1 &\\ 2A - 4B = -2 \end{array} \right.$

$\left\{ \begin{array}{l} A + B = 1 &\\ -A + 2B = 1 \end{array} \right.$

$3B = 2 \longrightarrow B = \frac{2}{3} \longrightarrow A = \frac{1}{3}$

$y = L^{-1} \{ \frac{\frac{1}{3}}{s - 4} + \frac{\frac{2}{3}}{s + 2} \} = \frac{1}{3} L^{-1} \{ \frac{1}{s - 4} \} + \frac{2}{3} L^{-1} \{ \frac{1}{s + 2} \}$

$y = \frac{1}{3} e^{4t} + \frac{2}{3} e^{-2t}$.

The System of Differential Equations

$\left\{ \begin{array}{l} y_1^{\prime} = 2 y_1 + 3 y_2 &\\ y_2^{\prime} = 4 y_1 - 2 y_2 \end{array} \right.$,

$y = f(x)$.

$y_1^{\prime \prime} = 2y_1^\prime + 3y_2^\prime \longrightarrow y_1^{\prime \prime} = 2y_1^\prime + 3(4y_1 - 2y_2) \longrightarrow y_1^{\prime \prime} = 2y_1^\prime + 12y_1 - 2(3y_2)$

$y_1^{\prime \prime} = 2y_1^\prime + 12y_1 - 2(y_1^\prime - 2y_1) \longrightarrow y_1^{\prime \prime} = 16y_1 \longrightarrow y_1^{\prime \prime} - 16y_1 = 0$

$r^2 - 16 = 0 \longrightarrow r^2 = 16 \longrightarrow r = \pm 4$

$\left\{ \begin{array}{l} y^{\prime \prime} + ay^\prime + by = 0 &\\ r^2 + ar + b = 0 \end{array} \right.$

$y_1 = c_1 e^{r_1 x} + c_2 e^{r_2 x} = c_1 e^{4x} + c_2 e^{-4x}$

$y_1^\prime = 4c_1 e^{4x} - 4c_2 e^{-4x} = 2(c_1 e^{4x} + c_2 e^{-4x}) + 3y_2$

$y_2 = \frac{1}{3} (4c_1 e^{4x} - 4c_2 e^{-4x} - 2(c_1 e^{4x} + c_2 e^{-4x}))$

$y_2 = \frac{1}{3} ((4c_1 - 2c_1) e^{4x} + (-2 c_2 - 4 c_2) e^{-4x})$

$y_2 = \frac{2}{3} c_1 e^{4x} - 2c_2e^{-4x}$

$\left\{ \begin{array}{l} y_1^\prime = y_1 + y_2 &\\ y_2^\prime = 4y_1 - 2y_2 \end{array} \right.$

$y = f(x)$

$y_1^{\prime \prime} = y_1^\prime + y_2^\prime \longrightarrow y_1^{\prime \prime} = y_1^\prime + (4y_1 - 2y_2)$

$y_1^{\prime \prime} = y_1^\prime + 4y_1 - 2(y_1^\prime - y_1) \longrightarrow y_1^{\prime \prime} = -y_1^\prime + 6y_1$

$y_1^{\prime \prime} + y_1^\prime - 6y_1 = 0$.

$\left\{ \begin{array}{l} r^2 + ar + b = 0 &\\ y^{\prime \prime} + ay^\prime + by = 0 \end{array} \right.$

$\left\{ \begin{array}{l} a = 1 &\\ b = -6 \end{array} \right.$

$r^2 + r - 6 = 0$

$r_{1, 2} = \frac{-1 \pm \sqrt{(1)^2 - 4 (1) (-6)}}{2} = \frac{-1 \pm \sqrt{1 - (-24)}}{2} = \frac{-1 \pm \sqrt{25}}{2}$

$\left\{ \begin{array}{l} r_1 = -3 &\\ r_2 = 2 \end{array} \right.$

$\Delta = 25 > 0$

$y_1 = c_1 e^{r_1 x} + c_2 e^{r_2 x} \longrightarrow y_1 = c_1 e^{2 x} + c_2 e^{-3 x}$

$y_1^\prime = 2 c_1 e^{2 x} - 3 c_2 e^{-3 x}$

$y_2 = y_1^\prime - y_1 = 2 c_1 e^{2 x} - 3 c_2 e^{-3 x} - y_1$

$y_2 = 2 c_1 e^{2 x} - 3 c_2 e^{-3 x} - c_1 e^{2 x} - c_2 e^{-3 x}$.

$\left\{ \begin{array}{l} y_1^\prime = 2y_1 - 5y_2 &\\ y_2^\prime = 5y_1 - 6y_2 \end{array} \right.$

$y = f(x)$

$y_1^{\prime \prime} = 2y_1^\prime - 5y_2^\prime = 2y_1^\prime - 5(5y_1 - 6y_2) = 2y_1^\prime - 25y_1 + 30y_2$

$-5y_2 = y_1^\prime - 2y_1 \longrightarrow y_2 = \frac{-1}{5}y_1^\prime + \frac{2}{5} y_1$

$y_1^{\prime \prime} = 2y_1^{\prime} - 25y_1 + 30(\frac{-1}{5}y_1^\prime + \frac{2}{5}y_1) = 2y_1^\prime - 25y_1 - 6y_1^\prime + 12y_1$

$y_1 ^{\prime \prime} = -4y_1^\prime - 13y_1 \longrightarrow y_1^{\prime \prime} + 4y_1^\prime + 13y_1 = 0$

$\left\{ \begin{array}{l} r^2 + ar + b = 0 &\\ y^{\prime \prime} + ay^\prime + by = 0 \end{array} \right.$

$\left\{ \begin{array}{l} a = 4 &\\ b = 13 \end{array} \right.$

$r^2 + 4r + 13 = 0$

$\sqrt{\Delta} = \sqrt{b_1^2 - 4a_1c_1} = \sqrt{(4)^2 - 4 (1) (13)} = \sqrt{16 - 52} = \sqrt{-36} = 6i$

$\Delta = -36 < 0$.

$\left\{ \begin{array}{l} r_1 = \alpha + i \beta &\\ r_2 = \alpha - i \beta \end{array} \right.$

$\left\{ \begin{array}{l} r_1 = \frac{-4 + 6i}{2} &\\ r_2 = \frac{-4 - 6i}{2} \end{array} \right.$

$\left\{ \begin{array}{l} r_1 = -2 + 3i &\\ r_2 = -2 - 3i \end{array} \right.$

$\left\{ \begin{array}{l} \alpha = -2 &\\ \beta = 3 \end{array} \right.$

$y = e^{\alpha x} (c_1 \ cos(\beta x) + c_2 \ sin(\beta x))$

$y_1 = e^{-2x} (c_1 \ cos(3x) + c_2 \ sin(3x))$

$y_1^\prime = -2e^{-2x}(c_1cos(3x) + c_2sin(3x)) + e^{-2x}(-3c_1sin(3x) + 3c_2cos(3x))$

$y_1^\prime = 2y_1 - 5y_2 \longrightarrow y_2 = \frac{-1}{5} y_1^\prime + \frac{2}{5}y_1$,

$y_2 = \frac{-1}{5}(-2e^{-2x}(c_1cos(3x) + c_2sin(3x)) + e^{-2x}(-3c_1sin(3x) + 3c_2cos(3x))) + \frac{2}{5} e^{-2x}(c_1cos(3x) + c_2sin(3x))$.

References