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Solve for y(t) with the initial condition y(0) = 0 (write the answer in explicit form): y'(t) = e^(5t - y).

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Final answer:

The differential equation y'(t) = e^(5t - y) with the initial condition y(0) = 0 is solved by separating variables, integrating, and then applying the initial condition to find the integration constant. The final explicit solution for y(t) is y = ln(1/5 * e^(5t) + 4/5).

Step-by-step explanation:

The task is to solve the differential equation y'(t) = e^(5t - y) with the initial condition y(0) = 0. This is a separable differential equation since we can isolate all the y terms on one side and all the t terms on the other side.

First, we rearrange the equation to separate variables:

\(\frac{dy}{dt} = e^{5t}e^{-y}\). Then, we take \(e^y\) to the other side to get

\(e^y dy = e^{5t} dt\). Now, we can integrate both sides:

\(\int e^y dy = \int e^{5t} dt\), leading to

\(e^y = \frac{1}{5}e^{5t} + C\).

Applying the initial condition y(0) = 0, we find that

\(e^0 = \frac{1}{5}e^0 + C\) which simplifies to

1 = \frac{1}{5} + C, giving us

C = \frac{4}{5}. Hence,

\(e^y = \frac{1}{5}e^{5t} + \frac{4}{5}\).

To solve for y explicitly, we take the natural logarithm of both sides:

\(y = ln\left(\frac{1}{5}e^{5t} + \frac{4}{5}\right)\). This is the solution in explicit form.

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