Ytukyg

Solve the differential equation Solve $$frac{d^2y}{dx^2}+2frac{dy}{dx}+y=x sin^2x$$

i used the substitution $frac{dy}{dx}+y=t$ we get

$$frac{dt}{dx}+t=xsin^2x$$ which is linear first order, using integrating factor method we get the solution

$$te^x=int xe^xsin^2xdx+C$$

$$te^x=frac{1}{2} left(int xe^xdx-int xe^xcos 2xdxright)+C=frac{1}{2}left(P-Qright)+C tag{1}$$

we have $$P=int xe^x=(x-1)e^x+C'$$

now using integration by parts we have

$$Q=int xe^x cos2xdx=xint e^xcos 2xdx-int left(int e^xcos 2xdxright)dx$$

$$int xe^x cos2xdx=frac{xe^x}{5}left(cos 2x+2sin 2xright)-frac{1}{5}int e^x(cos 2x+2sin 2x)dx $$

$$int xe^x cos2xdx=frac{xe^x}{5}left(cos 2x+2sin 2xright)-frac{1}{5}left(frac{e^x}{5}(cos 2x+2sin 2x)right)-left(frac{2e^x}{5}(sin 2x-2 cos 2x)right)$$

finally we need to solve another linear differential equation viz:

$$e^xleft(frac{dy}{dx}+yright)=frac{1}{2}left(P-Qright)+C$$

which is very tedious.

is there a good approach?

Hint

If you are as lazy as I am, considering $$frac{d^2y}{dx^2}+2frac{dy}{dx}+y=frac{x}{2}(1-cos(2x))$$ assume
$$y_p=a+b x+csin(2x)+dcos(2x)+e xsin(2x)+f xcos(2x)$$ Replace and identify $a,b,c,d,e,f$. This should be simple.

Write the differential equation as $$(D^2+2D+1)y=xsin^2(x)=frac{x}{2}(1-cos(2x))$$ then a complete solution can be written as $$y=y_r+y_p$$ where $y_r$ is the most general solution of the reduced equation $(D^2+2D+1)y=(D+1)^2y=0$ i.e. $y_r=ae^{-x}+bxe^{-x}$ , where $a,b$ are two arbitrary constants.

And $y_p$ is a one particular integral given by $$y_p=frac{1}{(D^2+2D+1)}frac{x}{2}(1-cos(2x))$$ i.e. $2y_p=(D+1)^{-2}(x)-(D+1)^{-2}(xcos(2x))$. Now $$(1+D)^{-2}(x)={1+frac{(-2)}{1!}D+frac{(-2)(-2-1)}{2!}+....}=(1-2D)x=x-2.$$ Since $D^2x=0,D^3x=0,....$

Next $(1+D)^{-2}(xcos(2x))=Re((1+D)^{-2}(xe^{iota 2x}))$. So that $$(1+D)^{-2}(xe^{iota 2x})=e^{iota 2x}frac{1}{1+2(D+2iota)+(D+2iota)^2} x.$$ Again expanding as bi-nomial series as in pervious part and then considering real part you can find particular integral $y_p$.

The homogeneous part is classical, though the characteristic polynomial has a double root, giving terms $e^{-x}$ and $xe^{-x}$.

The non-homogeneous one is a little harder, though you are lucky that there is no overlap with the roots. Let us try a solution of the form $y=xz+w$. The equation turns to

$$(z''+2z'+z)x+2z'+z+w''+2w'+w=xfrac{1-cos 2x}2=xfrac{1-Re(e^{i2x)}}2.$$

The constant term is dealt with by $dfrac12$, while the other can be handled using complex numbers, giving

$$z=-frac12Releft(frac{e^{i2x}}{4+2i+1}right).$$

Next,

$$w''+2w'+w=frac12Releft(frac{(i4+1))e^{i2x}}{4+2i+1}right)$$ and

$$w=frac12Releft(frac{(i4+1))e^{i2x}}{(4+2i+1)^2}right).$$