Question

a) Solve the differential equation
dy/dx + xy= y^2 e^(x^2/2) sinx

b) Given that
y1(x) = x^(-1) is one solution of the differential equation

2x^2 y ′′ + 3xy′ −y =0 ,x >0,
find a second linearly independent solution of the equation

Answer 1

a)

$\frac{dy}{dx}+xy=y^2e^{\frac{x^2}{2}}\sin x\\ y'+xy=y^2e^{\frac{x^2}{2}}\sin x$

We divide by $y^2$

$\frac{y'}{y^2}+\frac{x}{y}=e^{\frac{x^2}{2}}\sin x\\$

Replacement

$y^{-1}=z, y=z^{-1}, y'=-z^2\cdot z'\\ -\frac{z^{-2}z'}{z^{-2}}+xz=e^{\frac{x^2}{2}}\sin x\\ -z'+xz=e^{\frac{x^2}{2}}\sin x\\ i) -z'+xz=0\\ \frac{dz}{dx}=xz\\ \frac{dz}{z}=xdx\\ ln|z|=\frac{x^2}{2}+ln|c|\\ z_0=c\cdot e^{\frac{x^2}{2}}\\ ii) z=c(x)\cdot e^{\frac{x^2}{2}}\\$

Input $z$ in the equation $-z'+xz=e^{\frac{x^2}{2}}\sin x$

$-c'e^{\frac{x^2}{2}}-xe^{\frac{x^2}{2}}c+xe^{\frac{x^2}{2}}c=e^{\frac{x^2}{2}}\sin x\\ c'=-\sin x\\ c=\cos x+c_1$

then $z=(\cos x+c_1)e^{\frac{x^2}{2}}$

or $y^{-1}=(\cos x+c_1)e^{\frac{x^2}{2}}$

and solutoin of equation is function

$y=0$

b)

$y_1(x)=\frac{1}{x}\\ 2x^2y''+3xy'-y=0$

for equation

$a_0(x)y''+a_1(x)y'+a_2(x)y=0$

used formula Ostrohrad-Liouville

$\begin{vmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{vmatrix}=ce^{-\int \frac{a_1(x)}{a_0(x)}dx}$

in our case

$a_0(x)=2x^2, a_1(x)=3x, a_2(x)=-1$

then

$\begin{vmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{vmatrix}=ce^{-\int \frac{3x}{2x^2}dx}\\ y_1y'_2-y'_1y_2=ce^{- \frac{3}{2}\int\frac{1}{x}dx}\\ y_1y'_2-y'_1y_2=ce^{- \frac{3}{2}ln|x|}\\ y_1y'_2-y'_1y_2=cx^{- \frac{3}{2}}$

Divide by $y_1^2$

$\frac{y_1y'_2-y'_1y_2}{y_1^2}=cx^{- \frac{3}{2}}x^2\\ (\frac{y_2}{y_1})'=cx^{\frac{1}{2}}\\ \frac{y_2}{y_1}=\int cx^{\frac{1}{2}}dx\\ \frac{y_2}{y_1}=\frac{2}{3}cx^{\frac{3}{2}}+c_1$

Then the solution of equation is

$y=(\frac{2}{3}cx^{\frac{3}{2}}+c_1)\frac{1}{x}$

or

$y=\frac{2}{3}cx^{\frac{1}{2}}+c_1\frac{1}{x}$

So second linearly independent solution of the equation is

$y_2=\frac{2}{3}x^{\frac{1}{2}}$