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Solve the initial value problem. dt dy ​ =7e −t sec 2 (7πe −t ),y(ln4)=4/π y= (Type an exact answer, using π as needed.)

User Face
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Final answer:

The solution to the initial value problem
\( \tan(7\pi e^(-t)) = -7e^(-t) - (7)/(4) + \tan\left((7\pi)/(4)\right) \), with the constant determined by the initial condition
\( y(\ln 4) = (4)/(\pi) \).

Explanation:

To solve the initial value problem
\((dy)/(dt) = 7e^(-t)\sec^2(7\pi e^(-t))\) with the initial condition
\(y(\ln 4) = (4)/(\pi)\), we need to separate variables and integrate.

First, separate variables:


\[(1)/(\sec^2(7\pi e^(-t)))dy = 7e^(-t)dt\]

Integrate both sides:


\[\int (1)/(\sec^2(7\pi e^(-t)))dy = \int 7e^(-t)dt\]

The left side involves the integral of the secant squared function, which is
\(\tan(7\pi e^(-t))\):


\[\tan(7\pi e^(-t)) = -7e^(-t) + C\]

Now, apply the initial condition
\(y(\ln 4) = (4)/(\pi)\):


\[\tan(7\pi e^(-\ln 4)) = -7e^(-\ln 4) + C\]

Simplify the expression and solve for
\(C\):


\[\tan\left((7\pi)/(4)\right) = -(7)/(4) + C\]


\[C = -(7)/(4) + \tan\left((7\pi)/(4)\right)\]

Now, substitute
\(C\) back into the equation:


\[\tan(7\pi e^(-t)) = -7e^(-t) -(7)/(4) + \tan\left((7\pi)/(4)\right)\]

This is the implicit solution to the initial value problem. If you need an explicit solution for
\(y\), it may involve inverse trigonometric functions, and the exact form will depend on how you choose to express it.

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