Answer to Question #143645 in Differential Equations for Nikhil

Question

Solve, using the method of variation of parameter,the following differential equation
d^2y/dx^2-2dy/dz=e^x.sinx

Answer 1

Let us solve the equation:

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

The characteristic equation of the homogeneous equation is the following:

$k^2-2=0$

Therefore, $k_1=-\sqrt{2}, k_2=\sqrt{2}$ .

According to the method of variation of parameter, the differential equation has the following solution:

$y(x)=C_1(x)e^{-\sqrt{2}x}+ C_2(x)e^{\sqrt{2}x}$ .

For functions $C_1(x)$ and $C_2(x)$ we have the following system for their derivatives:

$\begin{cases} C_1'(x)e^{-\sqrt{2}x}+ C_2'(x)e^{\sqrt{2}x} =0\\ -\sqrt{2} C_1'(x)e^{-\sqrt{2}x}+\sqrt{2} C_2'(x)e^{\sqrt{2}x} =e^x\sin x \end{cases}$

$\begin{cases} C_1'(x)e^{-\sqrt{2}x}+ C_2'(x)e^{\sqrt{2}x} =0\\ -C_1'(x)e^{-\sqrt{2}x}+ C_2'(x)e^{\sqrt{2}x} =\frac{1}{\sqrt{2}}e^x\sin x \end{cases}$

$\begin{cases} C_1'(x)e^{-\sqrt{2}x}+ C_2'(x)e^{\sqrt{2}x} =0\\ 2C_2'(x)e^{\sqrt{2}x} =\frac{1}{\sqrt{2}}e^x\sin x \end{cases}$

$\begin{cases} C_1'(x)=- C_2'(x)e^{2\sqrt{2}x} =0\\ C_2'(x) =\frac{1}{2\sqrt{2}}e^{(1-\sqrt{2})x}\sin x \end{cases}$

$\begin{cases} C_1'(x)=- \frac{1}{2\sqrt{2}}e^{(1+\sqrt{2})x}\sin x \\ C_2'(x) =\frac{1}{2\sqrt{2}}e^{(1-\sqrt{2})x}\sin x \end{cases}$

In the sequel we use the well known formula:

$\int e^{ax}\sin (bx) dx =\frac{b^2e^{ax}(a\sin(bx)-b\cos(bx))}{a^2(a^2+b^2)}+C$

$C_1(x)=-\frac{1}{2\sqrt{2}}\frac{e^{(1+\sqrt{2})x}((1+\sqrt{2})\sin(x)-\cos(x))}{(1+\sqrt{2})^2((1+\sqrt{2})^2+1)}+C_1=-\frac{1}{2\sqrt{2}}\frac{e^{(1+\sqrt{2})x}((1+\sqrt{2})\sin(x)-\cos(x))}{(3+2\sqrt{2})(4+2\sqrt{2})}+C_1=$

$=-\frac{1}{2\sqrt{2}}\frac{e^{(1+\sqrt{2})x}((1+\sqrt{2})\sin(x)-\cos(x))}{20+14\sqrt{2}}+C_1 =-\frac{e^{(1+\sqrt{2})x}((1+\sqrt{2})\sin(x)-\cos(x))}{56+40\sqrt{2}}+C_1$

$C_2(x)=\frac{1}{2\sqrt{2}}\frac{e^{(1-\sqrt{2})x}((1-\sqrt{2})\sin(x)-\cos(x))}{(1-\sqrt{2})^2((1-\sqrt{2})^2+1)}+C_2= \frac{1}{2\sqrt{2}}\frac{e^{(1-\sqrt{2})x}((1-\sqrt{2})\sin(x)-\cos(x))}{(3-2\sqrt{2})(4-2\sqrt{2})}+C_2=$

$=\frac{1}{2\sqrt{2}}\frac{e^{(1-\sqrt{2})x}((1-\sqrt{2})\sin(x)-\cos(x))}{20-14\sqrt{2}}+C_2= \frac{e^{(1-\sqrt{2})x}((1-\sqrt{2})\sin(x)-\cos(x))}{-56+40\sqrt{2}}+C_2$

Consequently, the solution is the following:

$y(x)=(-\frac{e^{(1+\sqrt{2})x}((1+\sqrt{2})\sin(x)-\cos(x))}{56+40\sqrt{2}}+C_1)e^{-\sqrt{2}x}+ (\frac{e^{(1-\sqrt{2})x}((1-\sqrt{2})\sin(x)-\cos(x))}{-56+40\sqrt{2}}+C_2)e^{\sqrt{2}x}$