# General solution to differential equation

• gruba
You should check those two steps again and see where the mistake is.In summary, the conversation discusses finding the general solution to a non-homogeneous differential equation using the method of undetermined coefficients. The attempt at a solution involves finding the particular solution and solving a system of equations to determine the coefficients. However, there is a mistake in the solution step, resulting in incorrect coefficients. The speaker suggests double-checking the evaluation of y'' and the solution step to find the error.
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 $A=2/5,B=21/25,C=−1/5,D=−3/25$

Last edited:
gruba said:

## 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.

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.

## 1. What is a general solution to a differential equation?

A general solution to a differential equation is a function or set of functions that satisfies the equation for all possible values of the independent variable. It includes all possible solutions to the equation, and can be written in a general form using arbitrary constants.

## 2. How is a general solution different from a particular solution?

A particular solution is a specific solution to a differential equation that satisfies the equation for specific values of the independent variable and initial conditions. A general solution, on the other hand, includes all possible solutions to the equation and can be obtained by adding any particular solution to the general solution.

## 3. What is the process for finding a general solution to a differential equation?

The process for finding a general solution to a differential equation involves solving the equation using integration techniques, such as separation of variables or substitution, to obtain a general solution with arbitrary constants. These constants can then be determined by applying initial conditions or additional information.

## 4. Can a general solution to a differential equation be verified?

Yes, a general solution can be verified by substituting it into the original differential equation and checking if it satisfies the equation for all possible values of the independent variable. It is important to note that a general solution may also include extraneous solutions that do not satisfy the initial conditions.

## 5. Are there any limitations to using a general solution to a differential equation?

One limitation is that a general solution may not be unique, as it may include multiple solutions that satisfy the equation. Additionally, it may not be possible to obtain a general solution for all types of differential equations, especially for more complex equations that require advanced mathematical techniques.

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