Answer to Question #161325 in Differential Equations for Priya

Question

Find the general solution of the equation xdy/dx+4y=x³e^x using the method of variation of parameters.

Answer 1

Solution:

Given differential equation cannot be solved by variation of parameters currently as there is insufficient data, either solution of the homogeneous equation should have been given or 'x' should be removed from 'dy/dx'. So, we are going to solve this by the method of Integrating factor.

$x\dfrac{dy}{dx}+4y=x^3e^x$

Dividing both sides by $x$ ,

$\dfrac{dy}{dx}+\dfrac4xy=x^2e^x$ ...(i)

On comparing this with $\dfrac{dy}{dx}+Py=Q$ , we get

$P=\dfrac4x,\ Q=x^2e^x$

Now, integrating factor, $I.F.=e^{\int Pdx}$

$I.F.=e^{\int \frac4xdx}=e^{4\ln x}=e^{\ln x^4}=x^4$

On multiplying (i) by this I.F., we get,

$I.F.\times y=\int Q\times I.F. \ dx$

$\Rightarrow x^4\times y=\int x^2e^x\times x^4\ dx$

$\Rightarrow yx^4=\int x^6e^x\ dx$

Now we apply integration by parts on right side, taking $x^6$ as first function and $e^x$ as second. Also we need to perform this rule various times further.

$\Rightarrow yx^4=x^6e^x-\int 6x^5e^x\ dx \\ \Rightarrow yx^4=x^6e^x-6[x^5e^x-\int 5x^4e^x]\ dx \\ \Rightarrow yx^4=x^6e^x-6x^5e^x+30[x^4e^x-\int 4x^3e^x]\ dx$

$\\ \Rightarrow yx^4=x^6e^x-6x^5e^x+30x^4e^x-120[x^3e^x-\int 3x^2e^x]\ dx \\ \Rightarrow yx^4=x^6e^x-6x^5e^x+30x^4e^x-120x^3e^x+360[x^2e^x-\int 2xe^x]\ dx$

$\\ \Rightarrow yx^4=x^6e^x-6x^5e^x+30x^4e^x-120x^3e^x+360x^2e^x-720(xe^x- e^x)+C \\ \Rightarrow yx^4=x^6e^x-6x^5e^x+30x^4e^x-120x^3e^x+360x^2e^x-720xe^x+720 e^x+C$

$\Rightarrow y=\dfrac{x^6e^x-6x^5e^x+30x^4e^x-120x^3e^x+360x^2e^x-720xe^x+720 e^x+C}{x^4}$

$\Rightarrow y=x^2e^x-6xe^x+30e^x-\dfrac{120e^x}{x}+\dfrac{360e^x}{x^2}-\dfrac{720e^x}{x^3}+\dfrac{720 e^x}{x^4}+\dfrac{C}{x^4}$