Answer to Question #126296 in Differential Equations for jse

Question

Solve the given differential equation by undetermined coefficients.

y'' + y' + y = x sin x

Answer 1

We solve a second order differential equation of the type:

ay''+by'+cy=f(x)

where a=1

b=1

c=1

f(x)=xsinx

The solution will consist of common and particular solutions and have a type:

y=y_com+y_part

To find a common solution we solve second order differential equation y''+y'+y=0 and find roots of polynomial r²+r+1=0

Its roots  

$r_1={{-1+\sqrt{1^2-4\sdot1\sdot1}}\over{2\sdot1}}={{{-1+\sqrt{-3}}\over{2}}={{-1+\sqrt3i}\over{2}}}={-0.5+{\sqrt{3}i\over{2}}}$

$r_2={{-1-\sqrt{1^2-4\sdot1\sdot1}}\over{2\sdot1}}={{{-1-\sqrt{-3}}\over{2}}={{-1-\sqrt3i}\over{2}}}={-0.5-{\sqrt{3}i\over{2}}}$

​Roots are complex therefore the common solution will be $y_{com}=e^{-{x\over2}}(C_1cos{{\sqrt3\over{2}}x}+C_2sin{{\sqrt3\over{2}}x})$

​As in the right side of the differential equation f(x)=xsinx so particular solution will have a type:

y_part=Axsinx+Bsinx+Cxcosx+Dcosx

y'_part=(A-D)sinx+(B+C)cosx+Axcosx-Cxsinx

y''part=-(B+2C)sinx-Dcosx-Cxcosx-Axsinx

Now differentiate and plug it into the differential equation

-(B+2C)sinx-Dcosx-Cxcosx-Axsinx+(A-D)sinx+(B+C)cosx+Axcosx-Cxsinx+Axsinx+Bsinx+Cxcosx+Dcosx=xsinx

(B+C)cosx+(A-2C-D)sinx-Cxsinx+Axcosx=xsinx

We need to pick A so that we get the same function on both sides of the equal sign. This means that the coefficients of the sines and cosines must be equal. Or,

xcox: A=0

cosx: B+C=0

xsinx: -C=1

sinx: A-2C-D=0

We get A=0, B=1, C=-1, D=2

y_part=sinx-xcosx+2cosx

We get final solution

$y=e^{-{x\over2}}(C_1cos{{\sqrt3\over{2}}x}+C_2sin{{\sqrt3\over{2}}x})+sinx-xcosx+2cosx$