Question #179384

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


1
Expert's answer
2021-04-15T07:26:11-0400

Given: y+4y=cos2xy{}''+4y=cos^2x

Require to solve the given differential equation.

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

is m2+4=0m^2+4=0


Solve the equation m2+4=0m^2+4=0

Now m2+4=0m=±2im^2+4=0\Rightarrow m=\pm 2i


General solution to the homogeneous differential equation is

yh=c1cos2x+c2sin2xy_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 yp=Au+Bv,=cos2x,v=sin2xy_p=Au+Bv,=cos2x,v=sin2x and

A=vg(x)uvvudx,B=ug(x)uvvudxA=-\int \frac{vg(x)}{uv{}'-vu{}'}dx,B=\int \frac{ug(x)}{uv{}'-vu{}'}dx , where g(x)=cos2xg(x)=cos^2x


First let us find uvvuuv{}'-vu{}'


Now u=cos2x,v=sin2xuvvu=(cos2x)(2cos2x)(sin2x)(2sin2x)=2cos22x+2sin22x=2(1)=2u=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=vg(x)uvvudxA=(sin2x)(cos2x)2dx=(2sinxcosx)(cos2x)2dxA=-\int \frac{vg(x)}{uv{}'-vu{}'}dx\Rightarrow A=-\int \frac{(sin2x)(cos^2x)}{2}dx=-\int \frac{(2sinxcosx)(cos^2x)}{2}dx

A=t3(dt),t=cosxdt=sinxdx\Rightarrow A=-\int t^3(-dt),t=cosx\Rightarrow dt=-sinxdx

Then, we get A=t44=cos4x4A=-\frac{t^4}{4}=-\frac{cos^4x}{4}


And

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

B=12(cos2x)[1+cos2x2]dx\Rightarrow B=\frac{1}{2}\int(cos2x)\left [ \frac{1+cos2x}{2} \right ]dx

B=14(cos2x)dx+14cos22xdx\Rightarrow B=\frac{1}{4}\int(cos2x)dx+\frac{1}{4}\int cos^2{2x}dx

B=14(cos2x)dx+141+cos4x2dx\Rightarrow B=\frac{1}{4}\int(cos2x)dx+\frac{1}{4}\int \frac{1+cos4x}{2}dx

B=14[sin2x2]+18[x+sin4x4]\Rightarrow B=\frac{1}{4}\left [ \frac{sin2x}{2} \right ]+\frac{1}{8}\left [ x+\frac{sin4x}{4} \right ]

B=18sin2x+x8+sin4x32\Rightarrow B=\frac{1}{8}sin2x+\frac{x}{8}+\frac{sin4x}{32}


Using the above, we have

yp=14(cos4x)(cos2x)+[18sin2x+x8+sin4x32](sin2x)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=yh+ypy=y_h+y_p


y=c1cos2x+c2sin2x14(cos4x)(cos2x)+[18sin2x+x8+sin4x32](sin2x)y=c_1cos2x+c_2sin2x-\frac{1}{4}(cos^4x)(cos2x)+\left [ \frac{1}{8}sin2x+\frac{x}{8}+\frac{sin4x}{32} \right ](sin2x)


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