Find the solution to the initial value problemy′−e−t+5y=−5,   y(0)=−1.𝑦′−𝑒−𝑡+5𝑦=−5,   𝑦(0)=−1.

The given differential equation is a first order linear differential equation. The general form of such an equation is y′ + p(t)y = g(t). In this case, p(t) = 5 and g(t) = e^-t - 5.

Step 1: Find the integrating factor e^(∫p(t) dt) = e^(∫5 dt) = e^(5t).

Step 2: Multiply every term of the differential equation by the integrating factor to get e^(5t)y′ - e^(4t) + 5e^(5t)y = -5e^(5t).

Step 3: The left-hand side of this equation is now the derivative of (e^(5t)y) with respect to t. So, we have (e^(5t)y)' = -5e^(5t) + e^(4t).

Step 4: Integrate both sides with respect to t to find y. ∫(e^(5t)y)' dt = ∫(-5e^(5t) + e^(4t)) dt.

Step 5: This gives e^(5t)y = -e^(5t) + 1/5e^(4t) + C, where C is the constant of integration.

Step 6: Finally, solve for y to get y = -1 + 1/5e^(-t) + Ce^(-5t).

Step 7: Use the initial condition y(0) = -1 to find C. -1 = -1 + 1/5 + C, so C = -1/5.

Therefore, the solution to the initial value problem is y = -1 + 1/5e^(-t) - 1/5e^(-5t).