MHB Solve 2nd Order Linear Inhomogeneous ODE: Muhammad Fasih

MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$y(x)=y_h(x)+y_p(x)=c_1\cos(2x)+c_2\sin(2x)+\frac{5x-2}{25}e^x+\frac{x}{16}\left(\sin(2x)-2x\cos(2x) \right)$$
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top