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Clara I's question at Yahoo! Answers regarding verifying a given solution to an ODE

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MarkFL

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Feb 24, 2012
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Here is the question:

Differential Equations HW HELP!?

13. y'' + y = sec(t), 0< t < π/2; y = (cos t) ln cos t + t sin t

In each of Problems 7 through 14, verify that each given function is a solution of the differential equation.

I'm currently trying to review derivatives, but trig derivatives are super hard. 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 Clara I,

If the given solution is indeed a solution to the ODE, then we should find by adding the second derivative of the solution to itself, gives $\sec(t)$.

In order to differentiate the given solution with respect to $t$, we will need the following rules:

The Product Rule:

\(\displaystyle \frac{d}{du}\left(f(u)g(u) \right)=f(u)\frac{d}{du}(g(u))+\frac{d}{du}(f(u))g(u)\)

The Logarithmic Rule:

\(\displaystyle \frac{d}{du}(\ln(u))=\frac{1}{u}\)

The Chain Rule:

\(\displaystyle \frac{d}{du}\left(f(g(u)) \right)=\frac{df}{dg}\cdot\frac{dg}{du}\)

Trigonometric Rules:

\(\displaystyle \frac{d}{du}\left(\sin(u) \right)=\cos(u)\)

\(\displaystyle \frac{d}{du}\left(\cos(u) \right)=-\sin(u)\)

Using these rules, we may now compute:

\(\displaystyle y=\cos(t)\ln\left(\cos(t) \right)+t\sin(t)\)

\(\displaystyle \frac{dy}{dt}=\cos(t)\cdot\frac{-\sin(t)}{\cos(t)}-\sin(t)\ln\left(\cos(t) \right)+t\cos(t)+\sin(t)=t\cos(t)-\sin(t)\ln\left(\cos(t) \right)\)

\(\displaystyle \frac{d^2y}{dt^2}=-t\sin(t)+\cos(t)+\frac{\sin^2(t)}{\cos(t)}-\cos(t)\ln\left(\cos(t) \right)=-t\sin(t)+\frac{\cos^2(t)+\sin^2(t)}{\cos(t)}-\cos(t)\ln\left(\cos(t) \right)=\)

\(\displaystyle -t\sin(t)+\sec(t)-\cos(t)\ln\left(\cos(t) \right)\)

And so we find:

\(\displaystyle \frac{d^2y}{dt^2}+y=-t\sin(t)+\sec(t)-\cos(t)\ln\left(\cos(t) \right)+\cos(t)\ln\left(\cos(t) \right)+t\sin(t)=\sec(t)\)

Thus, we have shown the given solution does in fact satisfy the given ODE.