MHB Solve 2nd Order Linear Inhomogeneous ODE: Muhammad Fasih

AI Thread Summary
The discussion focuses on solving the second-order linear inhomogeneous ordinary differential equation (ODE) y'' + 4y = xe^x + xsin(2x) using the method of undetermined coefficients. The annihilator method is employed to determine the appropriate form of the particular solution, leading to the conclusion that the operator (D-1)^2(D^2+4)^2 annihilates the right-hand side of the equation. The general solution combines both the homogeneous and particular solutions, resulting in a comprehensive expression for y(x). The final solution is presented as a combination of exponential and trigonometric functions, reflecting the contributions from both the homogeneous and particular components. The thread effectively illustrates the application of advanced techniques in solving complex differential equations.
MarkFL
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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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Hello Muhammad Fasih,

We are given the following ODE to solve:

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

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:

$$f(x)=xe^x$$

Differentiating, we find:

$$f'(x)=xe^x+e^x$$

$$f''(x)=xe^x+2e^x$$

Now, observing:

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

We may conclude the differential operator:

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

annihilates $f$.

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

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

Differentiating, we find:

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

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

Now, observing:

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

We may conclude the differential operator:

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

annihilates $g$.

Hence, the operator:

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

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

Applying the operator to the ODE, we obtain:

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

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

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

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:

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

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

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

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

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

Equating coefficients, we obtain the system:

$$5c_1+2c_2=0$$

$$5c_2=1$$

$$2c_5+4c_7=0$$

$$8c_8=0$$

$$-4c_4+2c_8=0$$

$$-8c_5=1$$

From this, we obtain:

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

And so our particular solution is:

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

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

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

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