Answer to Question #179384 in Differential Equations for Montse

Question

y''+4y=cos^2(x)

Answer 1

Given: $y{}''+4y=cos^2x$

Require to solve the given differential equation.

Auxiliary equation corresponding to the given homogeneous differential equation $y{}''+4y=0$

is $m^2+4=0$

Solve the equation $m^2+4=0$

Now $m^2+4=0\Rightarrow m=\pm 2i$

General solution to the homogeneous differential equation is

$y_h=c_1cos2x+c_2sin2x$

Now let us find the particular solution.

Use the method of variation of parameters to find the particular solution.

Using the method of variation of parameters, Particular solution is $y_p=Au+Bv,=cos2x,v=sin2x$ and

$A=-\int \frac{vg(x)}{uv{}'-vu{}'}dx,B=\int \frac{ug(x)}{uv{}'-vu{}'}dx$ , where $g(x)=cos^2x$

First let us find $uv{}'-vu{}'$

Now $u=cos2x,v=sin2x\Rightarrow uv{}'-vu{}'=(cos2x)(2cos2x)-(sin2x)(-2sin2x)=2cos^2{2x}+2sin^2{2x}=2(1)=2$

Using the above, we get

$A=-\int \frac{vg(x)}{uv{}'-vu{}'}dx\Rightarrow A=-\int \frac{(sin2x)(cos^2x)}{2}dx=-\int \frac{(2sinxcosx)(cos^2x)}{2}dx$

$\Rightarrow A=-\int t^3(-dt),t=cosx\Rightarrow dt=-sinxdx$

Then, we get $A=-\frac{t^4}{4}=-\frac{cos^4x}{4}$

And

$B=\int \frac{ug(x)}{uv{}'-vu{}'}dx\Rightarrow B=\int \frac{(cos2x)(cos^2x)}{2}dx$

$\Rightarrow B=\frac{1}{2}\int(cos2x)\left [ \frac{1+cos2x}{2} \right ]dx$

$\Rightarrow B=\frac{1}{4}\int(cos2x)dx+\frac{1}{4}\int cos^2{2x}dx$

$\Rightarrow B=\frac{1}{4}\int(cos2x)dx+\frac{1}{4}\int \frac{1+cos4x}{2}dx$

$\Rightarrow B=\frac{1}{4}\left [ \frac{sin2x}{2} \right ]+\frac{1}{8}\left [ x+\frac{sin4x}{4} \right ]$

$\Rightarrow B=\frac{1}{8}sin2x+\frac{x}{8}+\frac{sin4x}{32}$

Using the above, we have

$y_p=-\frac{1}{4}(cos^4x)(cos2x)+\left [ \frac{1}{8}sin2x+\frac{x}{8}+\frac{sin4x}{32} \right ](sin2x)$

Therefore general solution is

$y=y_h+y_p$

$y=c_1cos2x+c_2sin2x-\frac{1}{4}(cos^4x)(cos2x)+\left [ \frac{1}{8}sin2x+\frac{x}{8}+\frac{sin4x}{32} \right ](sin2x)$