Question

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.)

Answer 1

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.