- Thread starter
- Admin
- #1
Muhammad Fasih's question at Yahoo! Answers regarding solving a 2nd order linear inhomogeneous ODE
- Thread starter MarkFL
- Start date
- Thread starter
- Admin
- #2
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:
\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1\cos(2x)+c_2\sin(2x)+\frac{5x-2}{25}e^x+\frac{x}{16}\left(\sin(2x)-2x\cos(2x) \right)\)
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:
\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1\cos(2x)+c_2\sin(2x)+\frac{5x-2}{25}e^x+\frac{x}{16}\left(\sin(2x)-2x\cos(2x) \right)\)