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Muhammad Fasih's question at Yahoo! Answers regarding solving a 2nd order linear inhomogeneous ODE

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  • #1


Staff member
Feb 24, 2012
Here is the question:

Solve the following differential equation y''+4y=xe^x+xsin2x by method of undetermined coefficients?

Please help me in doing this
I have no Idea of how to find the particular solution.
I have posted a link there to this thread so the OP may see my work.
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  • #2


Staff member
Feb 24, 2012
Hello Muhammad Fasih,

We are given the following ODE to solve:

\(\displaystyle y''+4y=xe^x+x\sin(2x)\)

Rather than referring to a table to determine the form of the particular solution, let's employ the annihilator method instead.

Let's look at the first term on the right side:

\(\displaystyle f(x)=xe^x\)

Differentiating, we find:

\(\displaystyle f'(x)=xe^x+e^x\)

\(\displaystyle f''(x)=xe^x+2e^x\)

Now, observing:

\(\displaystyle f''(x)-2f'(x)+f(x)=xe^x+2e^x-2\left(xe^x+e^x \right)+xe^x=0\)

We may conclude the differential operator:

\(\displaystyle A\equiv(D-1)^2\)

annihilates $f$.

Now, let's look at the second term on the right of the ODE.

\(\displaystyle g(x)=x\sin(2x)\)

Differentiating, we find:

\(\displaystyle g''(x)=4\left(\cos(2x)-x\sin(2x) \right)\)

\(\displaystyle g^{(4)}(x)=16\left(x\sin(2x)-2\cos(2x) \right)\)

Now, observing:

\(\displaystyle g^{(4)}(x)+8g''(x)+16g(x)=16\left(x\sin(2x)-2\cos(2x) \right)+32\left(\cos(2x)-x\sin(2x) \right)+16x\sin(2x)=0\)

We may conclude the differential operator:

\(\displaystyle A\equiv\left(D^2+4 \right)^2\)

annihilates $g$.

Hence, the operator:

\(\displaystyle C\equiv(D-1)^2\left(D^2+4 \right)^2\)

annhilates $f(x)+g(x)$.

Applying the operator to the ODE, we obtain:

\(\displaystyle \left((D-1)^2\left(D^2+4 \right)^3 \right)[y]=0\)

And for this homogeneous ODE with repeated characteristic roots, we obtain the general solution:

\(\displaystyle y(x)=\left(c_1+c_2x \right)e^x+\left(c_3+c_4x+c_5x^2 \right)\cos(2x)+\left(c_6+c_7x+c_8x^2 \right)\sin(2x)\)

Now, we know the solution to the original ODE will be the superposition of the homogenous and particular solutions. If we note that the homogenous solution is of the form:

\(\displaystyle y_h(x)=c_3\cos(2x)+c_6\sin(2x)\)

We may therefore conclude that the particular solution must have the form:

\(\displaystyle y_p(x)=\left(c_1+c_2x \right)e^x+\left(c_4x+c_5x^2 \right)\cos(2x)+\left(c_7x+c_8x^2 \right)\sin(2x)\)

Differentiating twice, and substituting into the original ODE, we obtain:

\(\displaystyle \left(5c_1+2c_2+5c_2x \right)e^x+\left(2c_5+4c_7+8c_8x \right)\cos(2x)+\left(-4c_4+2c_8-8c_5x \right)\sin(2x)=\left(0+1x \right)e^x+\left(0+0x \right)\cos(2x)+\left(0+1x \right)\sin(2x)\)

Equating coefficients, we obtain the system:

\(\displaystyle 5c_1+2c_2=0\)

\(\displaystyle 5c_2=1\)

\(\displaystyle 2c_5+4c_7=0\)

\(\displaystyle 8c_8=0\)

\(\displaystyle -4c_4+2c_8=0\)

\(\displaystyle -8c_5=1\)

From this, we obtain:

\(\displaystyle \left(c_1,c_2,c_4,c_5,c_7,c_8 \right)=\left(-\frac{2}{25},\frac{1}{5},0,-\frac{1}{8},\frac{1}{16},0 \right)\)

And so our particular solution is:

\(\displaystyle y_p(x)=\left(-\frac{2}{25}+\frac{1}{5}x \right)e^x+\left(-\frac{1}{8}x^2 \right)\cos(2x)+\left(\frac{1}{16}x \right)\sin(2x)\)

\(\displaystyle y_p(x)=\frac{5x-2}{25}e^x+\frac{x}{16}\left(\sin(2x)-2x\cos(2x) \right)\)

And so, the general solution to the given ODE is:

\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1\cos(2x)+c_2\sin(2x)+\frac{5x-2}{25}e^x+\frac{x}{16}\left(\sin(2x)-2x\cos(2x) \right)\)