Question

Solve the following initial value problem
d^2y/dx^2+dx/dy-2y=-6sin 2x-18cos2x
y(0)=2,y'(0)=2

Answer 1

$y''+y'-2y=-6 \sin 2x -18 \cos 2x$

Above equation is second order linear nonhomogeneous differential equation with constant coefficients. The general solution of this nonhomogeneous equation is the sum of the general solution $y_0 (x)$ of the related homogeneous equation ($y''+y'-2y=0$) and a particular solution $y_1(x)$ of the nonhomogeneous equation:

$y(x)=y_0(x)+y_1(x)$.

The characteristic equation is:

$k^2+k-2=0$

Roots of this equation is $k_1=1$ and $k_2=-2$, so general solution of the related homogeneous equation is $y_0 (x)=C_1e^x+C_2e^{-2x}$.

Let's use method of undetermined coefficient for obtaining of the particular solution of the nonhomogeneous equation $y_1(x)=A \sin 2x +B \cos 2x$.

Differentials:

$y_1'(x)=2A \cos 2x -2B \sin 2x$

$y_1''(x)=-4A \sin 2x -4B \cos 2x$

Substitute obtained functions into the initial equation:

$-4A \sin 2x -4B \cos 2x+2A \cos 2x -2B \sin 2x-2(A \sin 2x +B \cos 2x)=-6 \sin 2x -18 \cos 2x$

$-4A \sin 2x -4B \cos 2x+2A \cos 2x -2B \sin 2x-2A \sin 2x -2B \cos 2x=-6 \sin 2x -18 \cos 2x$

$\sin 2x (-4A-2B-2A)+ \cos 2x (-4B +2A -2B)=-6 \sin 2x -18 \cos 2x$

$\sin 2x (-6A-2B)+ \cos 2x (2A-6B)=-6 \sin 2x -18 \cos 2x$

$\begin{cases} -6A-2B=-6 \\ 2A-6B=-18 \end{cases}$

$\begin{cases} -6A-2B=-6 \\ 6A-18B=-54 \end{cases}$

$-20B=-60$

$B=3$

$2A-6 \cdot 3=-18$

$2A-18=-18$

$A=0$

So, $y_1(x)=3 \cos 2x$ and $y(x)=C_1e^x+C_2e^{-2x}+3 \cos 2x$.

According to the initial value problem:

$y'(x)=C_1e^x-2C_2e^{-2x}-6 \sin 2x$

$\begin{cases} C_1e^0+C_2e^0+3 \cos 0=2 \\ C_1e^0-2C_2e^0-6 \sin 0=2 \end{cases}$

$\begin{cases} C_1+C_2+3 =2 \\ C_1-2C_2=2 \end{cases}$

$\begin{cases} C_1+C_2 =-1 \\ C_1-2C_2=2 \end{cases}$

$3C_2=-3$

$C_2=-1$

$C_1-1 =-1$

$C_1=0$

Finally, $y(x)=-e^{-2x}+3 \cos 2x$