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Second order inhomogeneous differential equation

Vishak

New member
Oct 27, 2013
18
Hi MHB. I'm having yet another doubt regarding differential equations. Can someone please help me out? Thanks.

Consider the following differential equation:

\(\displaystyle {y}''+{y}'= x^{2}\)

I have found the homogeneous solution to be:

\(\displaystyle y_{H}=c_{1} + c_{2}e^{-x}\)

But when finding the particular solution, using reduction of order, I end up getting:

\(\displaystyle y_{P}=\frac{x^{3}}{3} + \frac{cx^{2}}{2} + dx + e\)

By substituting the results for \(\displaystyle {y}''\) and \(\displaystyle {y}'\) back into the original equation, I am able to obtain \(\displaystyle c = -2\) and \(\displaystyle d = 2\). But what do I do about \(\displaystyle e\)?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hi MHB. I'm having yet another doubt regarding differential equations. Can someone please help me out? Thanks.

Consider the following differential equation:

\(\displaystyle {y}''+{y}'= x^{2}\)

I have found the homogeneous solution to be:

\(\displaystyle y_{H}=c_{1} + c_{2}e^{-x}\)

But when finding the particular solution, using reduction of order, I end up getting:

\(\displaystyle y_{P}=\frac{x^{3}}{3} + \frac{cx^{2}}{2} + dx + e\)

By substituting the results for \(\displaystyle {y}''\) and \(\displaystyle {y}'\) back into the original equation, I am able to obtain \(\displaystyle c = -2\) and \(\displaystyle d = 2\). But what do I do about \(\displaystyle e\)?
You are actually using the method of undetermined coefficients, not reduction of order, to find the particular solution. Because the right had side is quadratic, you want to assume the following form for your particular solution:

\(\displaystyle y_p(x)=Ax^2+Bx+C\)

But...since you already have a constant in your homogeneous solution, you need to multiply by a natural number power of $x$ so that no term in your particular solution is a solution to the homogeneous equation, hence you want:

\(\displaystyle y_p(x)=x\left(Ax^2+Bx+C \right)=Ax^3+Bx^2+Cx\)

As an alternate approach to solving this ODE, you could consider the substitution:

\(\displaystyle v=y'\)

and you would obtain a linear first order ODE in $v$.