Solve the initial value problem 7 (t + 1) dy/dt - 5 y = 10t, for t > -1 with y(0

Question

Solve the initial value problem
7 (t + 1) dy/dt - 5 y = 10t,
for t > -1 with y(0) = 11.
Find the integrating factor, u(t) = and then find y(t) =

Explanation / Answer

Write in standard form: dy/dt + [-7/(10(t+1))] y = 21t/[10(t+1)] Multiply both sides by the integrating factor e^(?-7 dt / (10(t+1)) = e^((-7/10) ln(t+1)) = (t + 1)^(-7/10): (t + 1)^(-7/10) dy/dt + (-7/10) (t + 1)^(-17/10) y = (21/10)t (t + 1)^(-17/10) ==> (d/dt) [(t + 1)^(-7/10) * y] = (21/10)t (t + 1)^(-17/10), by the product rule. Since ? t (t + 1)^(-17/10) dt = ? (z - 1) z^(-17/10) dz, via z = t + 1 = ? (z^(-7/10) - z^(-17/10)) dz = (10/3) z^(3/10) + (10/7) z^(-7/10) + C, = (10/3) (t + 1)^(3/10) + (10/7) (t + 1)^(-7/10) + C integrating both sides of the DE yields (t + 1)^(-7/10) * y = (21/10) [(10/3) (t + 1)^(3/10) + (10/7) (t + 1)^(-7/10)] + C ==> y = [7(t + 1) + 3] + C(t + 1)^(7/10) ==> y = (7t + 10) + C(t + 1)^(7/10). Solve for C via y(0) = 13: 13 = 10 + C ==> C = 3. Thus, the solution is y = (7t + 10) + 3(t + 1)^(7/10).

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