MHB Verifying a Solution to an ODE: Differential Equations HW Help!

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The discussion focuses on verifying the solution to the differential equation y'' + y = sec(t) for the function y = (cos t) ln(cos t) + t sin t. The user seeks assistance with derivatives, particularly trigonometric ones, to confirm the solution. By applying the Product, Logarithmic, Chain, and Trigonometric rules, the second derivative of y is computed. The calculations demonstrate that adding the second derivative to the original function yields sec(t), confirming that the proposed solution satisfies the differential equation. The verification process is successfully completed, affirming the correctness of the solution.
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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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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:

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

The Logarithmic Rule:

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

The Chain Rule:

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

Trigonometric Rules:

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

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

Using these rules, we may now compute:

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

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

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

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

And so we find:

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

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