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Jimmy Mai's questions at Yahoo! Answers regarding second order linear inhomogeneous ODEs

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MarkFL

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Feb 24, 2012
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Here are the questions:

Differential Equation 3.6?

Express the solution of the given initial value problem as a sum of 2 oscillations. Throughout, primes denote derivatives with respect to time t. Graph the solution function x(t) in such a way that you can identify its period.

2. x"+4x=5sin3t x(0)=x'(0)=0

4. x"+25x=90cos4t x(0)=0 x'(0)=90

Answer is x(t)=3/2 sin2t-sin3t and

x(t)=2√(106) cos(5t-alpha) + 10cos4t with alpha=pi-inverse tan(9/5) = 2.0779

I have no idea how to do this type of problem. Thank you for your help.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Jimmy Mai,

2.) We are given the IVP:

\(\displaystyle x"+4x=5\sin(3t)\) where \(\displaystyle x(0)=x'(0)=0\)

The first thing we want to do is find the associated homogeneous solution \(\displaystyle x_h(t)\). We see that the characteristic equation is:

\(\displaystyle r^2+4=0\)

Thus, the characteristic roots are:

\(\displaystyle r=\pm2i\)

and so we may state:

\(\displaystyle x_h(t)=c_1\cos(2t)+c_2\sin(2t)\)

We may choose, using a linear combination identity, to write this solution as:

\(\displaystyle x_h(t)=c_1\sin\left(2t+c_2 \right)\)

Now, using the method of undetermined coefficients, we may look for a particular solution \(\displaystyle x_p(t)\) of the form:

\(\displaystyle x_p(t)=A\sin(3t+B)\)

Differentiating twice with respect to $t$, we find:

\(\displaystyle x_p''(t)=-9A\sin(3t+B)\)

Substituting the particular solution into the original ODE, we may now determine the parameters $A$ and $B$:

\(\displaystyle \left(-9A\sin(3t+B) \right)+4\left(A\sin(3t+B) \right)=5\sin(3t)\)

\(\displaystyle -5A\sin(3t+B)=5\sin(3t+0)\)

Hence, we see that:

\(\displaystyle A=-1,\,B=0\)

and so:

\(\displaystyle x_p(t)=-\sin(3t)\)

By superposition, we may state the general solution to the ODE as:

\(\displaystyle x(t)=x_h(t)+x_p(t)=c_1\sin\left(2t+c_2 \right)-\sin(3t)\)

Now, to determine the solution satisfying the initial values, we need to differentiate the general solution with respect to $t$ to obtain:

\(\displaystyle x'(t)=2c_1\cos\left(2t+c_2 \right)-3\cos(3t)\)

Now we may write:

\(\displaystyle x(0)=c_1\sin\left(c_2 \right)=0\)

\(\displaystyle x'(0)=2c_1\cos\left(c_2 \right)-3=0\)

From these equations, we may observe that $c_1\ne0$, and so we have:

\(\displaystyle c_1=\frac{3}{2},\,c_2=0\)

Thus, the solution satisfying the IVP is:

\(\displaystyle x(t)=\frac{3}{2}\sin(2t)-\sin(3t)\)

A plot of the solution shows its period is $T=2\pi$:

jimmymai1.jpg
 
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MarkFL

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Feb 24, 2012
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4.) We are given the IVP:

\(\displaystyle x"+25x=90\cos(4t)\) where \(\displaystyle x(0)=0,\,x'(0)=90\)

As before, we first want to find $x_h(t)$. We see the associated characteristic equation is:

\(\displaystyle r^2+5=0\)

and so:

\(\displaystyle r=\pm5i\)

Thus, the homogeneous solution may be written:

\(\displaystyle x_h(t)=c_1\cos\left(5t+c_2 \right)\)

Next, we may use undetermined coefficients to find a particular solution of the form:

\(\displaystyle x_p(t)=A\cos(4t+B)\)

Differentiating twice with respect to $t$, we find:

\(\displaystyle x_p''(t)=-16A\cos(4t+B)\)

Substituting into the original ODE, we find:

\(\displaystyle \left(-16A\cos(4t+B) \right)+25\left(A\cos(4t+B) \right)=90\cos(4t)\)

\(\displaystyle 9A\cos(4t+B)=90\cos(4t+0)\)

From this we determine:

\(\displaystyle A=10,\,B=0\)

and so we have:

\(\displaystyle x_p(t)=10\cos(4t)\)

By superposition, we may the give the general solution to the ODE as:

\(\displaystyle x(t)=x_h(t)+x_p(t)=c_1\cos\left(5t+c_2 \right)+10\cos(4t)\)

Now, to determine the solution satisfying the initial values, we need to differentiate the general solution with respect to $t$ to obtain:

\(\displaystyle x'(t)=-5c_1\sin\left(5t+c_2 \right)-40\sin(4t)\)

Now we may write:

\(\displaystyle x(0)=c_1\cos\left(c_2 \right)+10=0\)

\(\displaystyle x'(0)=-5c_1\sin\left(c_2 \right)=90\)

Solving both for $c_1$, we find:

\(\displaystyle c_1=-\frac{10}{\cos\left(c_2 \right)}=-\frac{18}{\sin\left(c_2 \right)}\)

\(\displaystyle 5\sin\left(c_2 \right)=9\cos\left(c_2 \right)\)

\(\displaystyle \tan\left(c_2 \right)=\frac{9}{5}\)

\(\displaystyle c_2=\tan^{-1}\left(\frac{9}{5} \right)\)

\(\displaystyle c_1=-\frac{10}{\cos\left(\tan^{-1}\left(\frac{9}{5} \right) \right)}=-2\sqrt{106}\)

Thus we may state the solution satisfying the given IVP is:

\(\displaystyle x(t)=-2\sqrt{106}\cos\left(5t+\tan^{-1}\left(\frac{9}{5} \right) \right)+10\cos(4t)\)

Using the identity \(\displaystyle \cos(\theta-\pi)=-\cos(\theta)\), we may write this solution as:

\(\displaystyle x(t)=2\sqrt{106}\cos\left(5t-\left(\pi-\tan^{-1}\left(\frac{9}{5} \right) \right) \right)+10\cos(4t)\)

A plot of the solution shows its period is $T=2\pi$:

jimmymai2.jpg