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Find the solution of the following IVP. dy/dt -2ty = 9t^2 e^t^3, y(0) = -3. y(t) = .

User Kicsi
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Final answer:

To solve the given initial value problem (IVP), we can use the method of integrating factors. The integrating factor is e^(t^2), which is obtained by multiplying both sides of the given equation by e^(t^2). The particular solution is e^(t^2)y = ∫(9t^2e^(2t^3) dt) - 3.

Step-by-step explanation:

To solve the given initial value problem (IVP), we can use the method of integrating factors. The integrating factor is e^(t^2), which is obtained by multiplying both sides of the given equation by e^(t^2).

Multiplying the equation dy/dt - 2ty = 9t^2e^(t^3) by e^(t^2) gives us e^(t^2)dy/dt - 2tye^(t^2) = 9t^2e^(t^3)e^(t^2).

Now we can rewrite the left side of the equation as d/dt(e^(t^2)y) using the product rule. This gives us d/dt(e^(t^2)y) = 9t^2e^(2t^3).

Integrating both sides with respect to t gives us the general solution, which is e^(t^2)y = ∫(9t^2e^(2t^3) dt) + C, where C is the constant of integration.

To find the particular solution, we can use the initial condition y(0) = -3. Substituting t = 0 and y = -3 into the general solution equation, we get C = -3.

Therefore, the particular solution is e^(t^2)y = ∫(9t^2e^(2t^3) dt) - 3. To find the solution for y(t), we can divide both sides by e^(t^2).

User Degger
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