Answer to Question #105028 in Differential Equations for Vaishali

Question

Apply the method of variations of parameters to solve the following differential equations:
1) x^2 y" + xy' - y = x^2 e^x
2) y" + a^2 y = cosec ax
3) solve the equation d^2 y/dx^2 - cotx dy/dx - sin^2 xy = cosx - cos^3 x by changing the independent variable

Answer 1

1) $x^2y''+xy'-y=x^2e^x$

We will replace

$x=e^t, t=ln|x|\\ y'=\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{y'_t}{e^t}=e^{-t}y'_t\\ y''=\frac{dy'}{dx}=\frac{(e^{-t}\cdot y'_{t})'}{e^t}=\frac{y''_{t}e^{-t}-e^{-t}y'_t}{e^t}=e^{-2t}(y''_t-y'_t)$

then

$e^{2t}e^{-2t}(y''-y')+e^te^{-t}y'-y=e^{2t}\cdot e^{e^t}\\ y''-y'+y'-y=e^{2t}\cdot e^{e^t}\\ y''-y=e^{2t}\cdot e^{e^t}$

a)

$y''-y=0\\ \lambda^2-1=0\\ \lambda_1=1, \lambda_2=-1\\ y_1=e^t, y_2=e^{-t}\\ y_0=c_1e^t+c_2e^{-t}$

b)

$y=c_1(t)e^t+c_2(t)e^{-t}\\$

$\left\{\begin{matrix} c'_1e^t+c'_2e^{-t}=0 \\ c'_1e^t-c'_2e^{-t}=e^{2t}e^{e^t} \end{matrix}\right.\\ 2c'_1e^t=e^{2t}e^{e^t}\\ c'_1=\frac{1}{2}e^{t}e^{e^t}\\ c_1=\frac{1}{2}\int e^{t}e^{e^t}dt=\frac{1}{2}\int e^{e^t}d(e^t)=\\ =\frac{1}{2}e^{e^t}+k_1$

$2c'_2e^{-t}=-e^{2t}e^{e^t}\\ c'_2=-\frac{1}{2}e^{3t}e^{e^t}\\ c_2=-\frac{1}{2}\int e^{3t}e^{e^t}dt=-\frac{1}{2}\int e^{2t}e^{e^t}d(e^t)$

let $u=e^t$ then

$\int e^{2t}e^{e^t}de^t=\int u^2e^udu=u^2e^u-\int2ue^udu=\\ =u^2e^u-2(ue^u-\int e^udu)=u^2e^u-2ue^u+2e^u\\ c_2=-\frac{1}{2}(u^2e^u-2ue^u+2e^u)+k_2=\\ =-\frac{1}{2}(e^{2t}e^{e^t}-2e^te^{e^t}+2e^{e^t})+k_2$

Then

$y(t)=(\frac{1}{2}e^{e^t}+k_1)e^t+\\ +(-\frac{1}{2}(e^{2t}e^{e^t}-2e^te^{e^t}+2e^{e^t})+k_2)e^{-t}\\ y(x)=(\frac{1}{2}e^x+k_1)x+\\ +(-\frac{1}{2}(x^2e^x-2xe^x+2e^x)+k_2)\frac{1}{x}\\$

2)

$y''+a^2y=cosecax=\frac{1}{sinax}$

a)

$y''+a^2y=0\\ \lambda^2+a^2=0\\ \lambda_1=ai, \lambda_2=-ai\\ y_1=cosax, y_2=sinax\\ y_0=c_1cosax+c_2sinax$

b)

$y=c_{1}(x)cosax+c_2(x)sinax\\ \left\{\begin{matrix} c'_1cosax+c'_2 sinax=0 \\ -a c'_1sinax+ac'_2 cosax=\frac{1}{sinax} \end{matrix}\right.$

$\Delta=\begin{vmatrix} cos ax & sinax\\ -asinax & acosax \end{vmatrix}=a\\ \Delta_1=\begin{vmatrix} 0 & sinax\\ \frac{1}{sinax} & acosax \end{vmatrix}=-1\\ \Delta_2=\begin{vmatrix} cos ax & 0\\ -asinax & \frac{1}{sinax} \end{vmatrix}=\frac{cosax}{sinax}\\ c'_{1}=\frac{\Delta_1}{\Delta}=\frac{-1}{a}\\ c_{1}=\int\frac{-1}{a}dx=-\frac{x}{a}+k_1\\ c'_2=\frac{\Delta_2}{\Delta}=\frac{1}{a}\cdot\frac{cosax}{sinax}\\ c_2=\int\frac{1}{a}\cdot\frac{cosax}{sinax}dx=\frac{1}{a^2}ln|sinax|+k_2$

$y=(-\frac{x}{a}+k_1)cosax+(\frac{1}{a^2}ln|sinax|+k_2)sinax$

3) $y''-\cot x y'- \sin^2x y=\cos x-\cos^3x$

We will replace

$\cos x=t, x=\arccos t, \sin x =\sqrt{1-t^2}\\ y'=\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{y'_t}{-\frac{1}{\sqrt{1-t^2}}}=-\sqrt{1-t^2} y'_t\\ y''=\frac{dy'}{dx}=\frac{-y''_t\sqrt{1-t^2}-y'_t\frac{1}{2\sqrt{1-t^2}}(-2t)}{-\frac{1}{\sqrt{1-t^2}}}=\\ =y''_t(1-t^2)-2ty'_t$

input in equation

$y''(1-t^2)-ty'-\frac{t}{\sqrt{1-t^2}}(-\sqrt{1-t^2})y'-\\ -(1-t^2)y=t-t^3\\ y''(1-t^2)-(1-t^2)y=t(1-t^2)\\ y''-y=t,\\ y=y_0+y_1\\ a) y''-y=0\\ \lambda^2-1=0, \lambda_1=1, \lambda_2=-1\\ y_1=e^t, y_2=e^{-t}\\ y_0=c_1e^t+c_2e^{-t}\\ b) y_1=at+b$

input in equation

$-at-b=t\implies a=-1, b=0,\\ y_1=-t\\ y=c_1e^t+c_2e^{-t}-t\\ y=c_1e^{\cos x}+c_2e^{-\cos x}-\cos x$