Answer to Question #244374 in Differential Equations for Rohit

Question

Solve by the method of variation of parameters y"-y=xe^x/2

Answer 1

Solution:

$y^{\prime \prime}-y=\underbrace{x e^{x/ 2}}_{g(x)} \text {. }$

The auxiliary equation is given by:

$\begin{aligned} & D^{2}-1=0 \\ &\Rightarrow D^{2}=1 \Rightarrow \quad D=\pm 1 \end{aligned}$

complementary equation:$\quad y_{c}=c_{1} \underbrace{e^{ x}}_{y_{1}}+c_{2} \underbrace{e^{- x}}_{y_{2}}$

By variation of parameters,

Wronskian of the 2 functions is:

$W=\left|\begin{array}{ll} y_{1} & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{array}\right|=\left|\begin{array}{ll} e^{x} & e^{-{x}} \\ e^{x} & - e^{- x} \end{array}\right|$

$=-e^0-e^0 \\=-1-1 \\=-2$

Now, the particular solution is given by:

$\begin{aligned} y_{p}(x) &=y_{1} \int \frac{y_{2} g(x)}{W} d x+y_{2} \int \frac{y_{1} g(x)}{W} d x . \\ &=e^{{x}} \int \frac{e^{-{x}} \cdot x e^{{x/2}} d x}{-2}+e^{-{x}} \int \frac{e^{{x}} \cdot x e^{x / 2}}{-2} d x \\ \end{aligned}$

$=-\dfrac12 \int xe^{x / 2} d x-\dfrac12 \int xe^{x / 2} d x \\ =-\dfrac12 \int xe^{x / 2} d x\\$

On integrating by parts,

$y_p(x)=-\dfrac12[4\left(\frac{1}{2}e^{\frac{x}{2}}x-e^{\frac{x}{2}}\right)] \\=-2\left(e^{\frac{x}{2}}x-e^{\frac{x}{2}}\right) \\=2e^\frac{x}{2}-2xe^\frac{x}{2}$

$\therefore$ The general solution is given by:

$y=y_{c}+y_{p} \\=c_{1} e^{{x}}+c_{2} e^{-{x}}+2e^\frac{x}{2}-2xe^\frac{x}{2}$