Solve the initial-value problem:
dv/dt -2tv = (3t²)(e^t²) , v(0) = 5
Thanks in advance!
dv/dt -2tv = (3t²)(e^t²) , v(0) = 5
Thanks in advance!
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The equation is first order linear. Use the integrating factor
exp(∫ -2t dt) = e^(-t²).
Multiply through and the left side is a product rule:
d/dt [e^(-t²) v] = 3t².
Integrate
e^(-t²) v = t^3 + C ==> v(t) = t^3 e^(t²) + C e^(t²).
Apply the initial condition to get 5 = 0 + C ==> C = 5. So
v(t) = (t^3+ 5)e^(t²).
exp(∫ -2t dt) = e^(-t²).
Multiply through and the left side is a product rule:
d/dt [e^(-t²) v] = 3t².
Integrate
e^(-t²) v = t^3 + C ==> v(t) = t^3 e^(t²) + C e^(t²).
Apply the initial condition to get 5 = 0 + C ==> C = 5. So
v(t) = (t^3+ 5)e^(t²).