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The resquest of the y'+6y=e^4t, y(0)=2

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This is a linear differential equation of first order. Solve this by integrating the coefficient of the y term and then raising e to the integrated coefficient to find the integrating factor, i.e. the integrating factor for this problem is e^(6x).
Multiplying both sides of the equation by the integrating factor:

(y')e^(6x) + 6ye^(6x) = e^(12x)

The left side is the derivative of ye^(6x), hence

d/dx[ye^(6x)] = e^(12x)

Integrating

ye^(6x) = (1/12)e^(12x) + c where c is a constant

y = (1/12)e^(6x) + ce^(-6x)

Use the initial condition y(0)=-8 to find c:

-8 = (1/12) + c
c=-97/12

Hence

y = (1/12)e^(6x) - (97/12)e^(-6x)
User Joe Saunders
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