Question

(d^2y/dx2)-2(dy/dx)+2y=sin2x

Answer 1

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

Let us solve a characteristic equation:

$k^2-2k+2=0$

$(k-1)^2+1=0$

$(k-1)^2=-1$

$k_1=1+i, k_2=1-i$

The general solution is of the following form:

$y(x)=e^x(C_1\cos x+C_2\sin x)+y_p(x),$ where $y_p(x)$ is a particular solution, $C_1, C_2$ are arbitrary constants.

The particular solution has the following form:

$y_p(x)=A\sin 2x+B\cos 2x$, whrere $A$ and $B$ are constants.

$\frac{dy_p}{dx}=2A\cos 2x-2B\sin 2x$

$\frac{d^2y_p}{dx^2}=-4A\sin 2x-4B\cos 2x$

Put founded derivatives in the differential equation:

$-4A\sin 2x-4B\cos 2x-2(2A\cos 2x-2B\sin 2x)+2(A\sin 2x+B\cos 2x)=\sin 2x$

$(-4A+4B+2A)\sin 2x+(-4B-4A+2B)\cos 2x=\sin 2x$

$(-2A+4B)\sin 2x+(-4A-2B)\cos 2x=\sin 2x$

Equating the corresponding coefficients, we obtain a system of equations:

$\begin{cases} -2A+4B=1\\ -4A-2B=0 \end{cases}$

$\begin{cases} -2A+4B=1\\ B=-2A \end{cases}$

$\begin{cases} -10A=1\\ B=-2A \end{cases}$

$\begin{cases} A=-\frac{1}{10}\\ B=\frac{1}{5} \end{cases}$

Finally, the solution of the differential equation is the following:

$y(x)=e^x(C_1\cos x+C_2\sin x)-\frac{1}{10}\sin 2x+\frac{1}{5}\cos 2x.$