General solution to differential equation

[gruba]

Homework Statement

Find the general solution to differential equation $y''-6y'+9y=e^x((2x+1)\cos x+(x+3)\sin x)$

Homework Equations

-Non homogeneous differential equation
-Homogeneous differential equation with constant coefficients
-Method of undetermined coefficients

The Attempt at a Solution

First, we find the solution for homogeneous equation $y''-6y'+9y=0\Rightarrow (\lambda -3)^2=0\Rightarrow \lambda_1=\lambda_2=3$.
Roots are real and multiple $\Rightarrow y_h=c_1e^{3x}+c_2xe^{3x}$.

We can find a particular solution for the equation by method of undetermined coefficients:

$$y_p=e^x((Ax+B)\cos x+(Cx+D)\sin x)=Axe^x\cos x+Be^x\cos x+Cxe^x\sin x+De^x\sin x$$
$${y'}_p=xe^x\cos x(A+C)+e^x\cos x(A+B+D)+xe^x\sin x(C-A)+(e^x\sin x(D+C-B))$$
$$y′′p=2Ce^x\cos x+e^x\cos x(2D+2C+2A)−2Axe^x\sin x+e^x\sin x(2C−2B−2A)$$

$$y′′_p−6y′_p+9y_p=xe^x\cos x(4C+3A)+e^x\cos x(−4A+3B+2C−4D)+xe^x\sin x(4A+3C)+e^x\sin x(−2A+4B−4C+3D)=2xe^x\cos x+e^x\cos x+xe^x\sin x+3e^x\sin x$$

Solving the system:
$4C+3A=2$
$−4A+3B+2C−4D=1$
$4A+3C=1$
$−2A+4B−4C+3D=3$

gives $A=−2/7,B=23/35,C=5/7,D=31/35$
that are wrong results.

Could someone check for possible errors?
Coefficients should be $A=2/5,B=21/25,C=−1/5,D=−3/25$

 

[Ray Vickson]

Homework Statement

Find the general solution to differential equation $y''-6y'+9y=e^x((2x+1)\cos x+(x+3)\sin x)$

Homework Equations

-Non homogeneous differential equation
-Homogeneous differential equation with constant coefficients
-Method of undetermined coefficients

The Attempt at a Solution

First, we find the solution for homogeneous equation $y''-6y'+9y=0\Rightarrow (\lambda -3)^2=0\Rightarrow \lambda_1=\lambda_2=3$.
Roots are real and multiple $\Rightarrow y_h=c_1e^{3x}+c_2xe^{3x}$.

We can find a particular solution for the equation by method of undetermined coefficients:

$$y_p=e^x((Ax+B)\cos x+(Cx+D)\sin x)=Axe^x\cos x+Be^x\cos x+Cxe^x\sin x+De^x\sin x$$
$${y'}_p=xe^x\cos x(A+C)+e^x\cos x(A+B+D)+xe^x\sin x(C-A)+(e^x\sin x(D+C-B))$$
$$y′′p=2Ce^x\cos x+e^x\cos x(2D+2C+2A)−2Axe^x\sin x+e^x\sin x(2C−2B−2A)$$

$$y′′_p−6y′_p+9y_p=xe^x\cos x(4C+3A)+e^x\cos x(−4A+3B+2C−4D)+xe^x\sin x(4A+3C)+e^x\sin x(−2A+4B−4C+3D)=2xe^x\cos x+e^x\cos x+xe^x\sin x+3e^x\sin x$$

Solving the system:
$4C+3A=2$
$−4A+3B+2C−4D=1$
$4A+3C=1$
$−2A+4B−4C+3D=3$

gives $A=−2/7,B=23/35,C=5/7,D=31/35$
that are wrong results.

Could someone check for possible errors?
Coefficients should be $A=2/5,B=21/25,C=−1/5,D=−3/25$

You can check this for yourself (and you should always do that): just substitute your "solution" into the DE, to see if it works.

 

[RUber]

I checked your evaluation for y', and it matches mine. I was able to get the answer you put as the solution, so your mistake must be in either the evaluation of y'' or the solution step.