Find the solution of the following IVP: $$ \frac{d y}{d t}-2 t y=-6 t^{2} e^{t^{2}}, \quad y(0)=-1 $$ $$ y(t)= $$

Question

UpStudy AI · Accepted Answer

We are given the initial value problem

$$
\frac{d y}{d t} - 2 t y = -6 t^{2} e^{t^{2}}, \quad y(0) = -1.
$$

**Step 1. Identify the integrating factor**

The differential equation is linear and in the form

$$
y' - 2ty = -6t^2e^{t^2}.
$$

The integrating factor, $$\mu(t)$$, is given by

$$
\mu(t) = e^{\int -2t\, dt} = e^{-t^2}.
$$

**Step 2. Multiply the differential equation by the integrating factor**

Multiply both sides of the equation by $$e^{-t^2}$$:

$$
e^{-t^2} y' - 2t e^{-t^2} y = -6t^2.
$$

The left-hand side is recognized as the derivative of $$\left(e^{-t^2}y\right)$$:

$$
\frac{d}{dt}\left(e^{-t^2} y\right) = -6t^2.
$$

**Step 3. Integrate both sides**

Integrate with respect to $$t$$:

$$
\int \frac{d}{dt}\left(e^{-t^2} y\right)dt = \int -6t^2\, dt.
$$

This gives

$$
e^{-t^2} y = -6 \int t^2\, dt = -6 \cdot \frac{t^3}{3} + C = -2t^3 + C,
$$

where $$C$$ is the constant of integration.

**Step 4. Solve for $$y(t)$$**

Multiply both sides by $$e^{t^2}$$ to isolate $$y$$:

$$
y(t) = e^{t^2} \left(-2t^3 + C\right).
$$

**Step 5. Apply the initial condition**

We are given $$y(0) = -1$$. Substitute $$t=0$$ into the solution:

$$
y(0) = e^{0} \left(-2 \cdot 0^3 + C\right) = C = -1.
$$

Thus, $$C = -1$$.

**Final Answer**

Substitute $$C$$ back into the solution:

$$
y(t) = e^{t^2} \left(-2t^3 - 1\right).
$$
$$ y(t) = e^{t^2} (-2t^3 - 1) $$

Find the solution of the following IVP: