Question

find the solution of the given initial value problem. $$ y^{\prime}-y=2 t e^{2 t}, \quad y(0)=1 $$

Short answer

Question: Given the differential equation \(y' - y = 2te^{2t}\) with initial condition \(y(0) = 1\), find the solution y(t). Answer: The solution to the initial value problem is \(y(t) = (2t-2)e^{2t} + 3e^{t}\).

Step-by-step solution

Rewrite as a first-order linear equation in standard form

We can notice that the given differential equation is already in standard form: $$ y^{\prime}-y=2 t e^{2 t} $$ Here, the function p(t) is -1, and the function g(t) is 2te^{2t}.

Determine the integrating factor

We'll now find the integrating factor \(μ(t)\) by calculating the exponential of the integral of p(t) with respect to t: $$ μ(t) = e^{\int -1 dt} = e^{-t} $$

Multiply by the integrating factor

Next, multiply both sides of the equation by the integrating factor, \(e^{-t}\): $$ e^{-t} (y^{\prime}-y)=2 t e^{t} $$

Integrate both sides with respect to t

Now, we'll find that the left side of the equation is the derivative of the product \(y(t)e^{-t}\). Let's integrate both sides with respect to t: $$ \int y(t) e^{-t} dt = \int 2 t e^{t} dt $$

Solve the integral on the right side

To solve the integral on the right side, we will use integration by parts: $$ \int 2 t e^{t} dt = (2t-2)e^{t} + C $$

Integrate left side and solve for y(t)

Now the equation looks like: $$ y(t) e^{-t} = (2t-2)e^{t} + C $$ Let's multiply both sides by \(e^{t}\) to solve for y(t): $$ y(t) = (2t-2)e^{2t} + Ce^{t} $$

Apply the initial condition and find C

Lastly, we'll apply the initial condition y(0) = 1 to find C: $$ 1 = (2(0)-2)e^{2(0)} + Ce^{(0)} $$ This simplifies to: $$ 1 = -2 + C $$ So, C = 3.

Write the final solution

Now, we have found the solution to the initial value problem: $$ y(t) = (2t-2)e^{2t} + 3e^{t} $$