Question

A first-order equation Consider the equation t² y′(t) + 2ty(t) = e⁻ᵗc. Find the solution that satisfies the condition y(1) = 0

Accepted Answer

Rewrite the given differential equation in standard linear form. The original equation is $$t^{2} y$$'(t) + 2t y(t) = $$e^{-t}. $$Divide both sides by $$t^{2} ($$assuming $$t 
eq 0) to $$get $$y'(t) + \frac{2}{t} y(t) = \frac{e$$^{-t}$$}{t$$^{2}$$}. $$Identify the integrating factor (IF) for the linear first-order ODE. The integrating factor is given by $$\mu(t) = e$$^{\int P(t) dt}$$, where P(t$$) = \frac{2}{t}$$. Calculate $$\mu(t) = $$e^{\int \frac{2}$${t} dt} = $$e^{2 \ln |t|}$$ = $$t^{2}. $$Multiply the entire differential equation by the integrating factor $$t^{2} to $$write the left side as a derivative of a product: $$t^{2} y$$'(t) + 2t y(t) = \frac{d}{dt} [$$t^{2} y(t$$)]$$. So, the equation becomes $$\frac{d}{dt} [$$t^{2} y(t$$)] = $$e^{-t}. $$Integrate both sides with respect to $$t$$: $$\int \frac{d}{dt} [t$$^{2}$$ y(t)] dt = \int e$$^{-t}$$ dt. $$This gives $$t^{2} y(t$$) = $$-e^{-t} + C$, where C$ is the constant of integration. Apply the initial condition y(1) = 0$ to find C$. Substitute t=1$ and y(1)=0$ into t$$^{2}$$ y(t) = -e$$^{-t}$$ + C to $$get $$1^{2}$$ \cdot 0 = $$-e^{-1} + C$, which simplifies to C$$ = $$e^{-1}. $$Then express $$y(t)$$ explicitly as $$y(t) = \frac{-e$$^{-t}$$ + e$$^{-1}$$}{t$$^{2}$$}.$$

A first-order equation Consider the equation t² y′(t) + 2ty(t) = e⁻ᵗ
c. Find the solution that satisfies the condition y(1) = 0

Verified video answer for a similar problem:

Key Concepts

First-Order Linear Differential Equations

Integrating Factor Method

Initial Value Problems