Question

Find a solution to y'(t) = te^-t satisfying the condition y(1) = 1.

Answer 1

Answer:

$y=-e^{-t}(t+1)+1+\frac{2}{e}$

Step-by-step explanation:

The given differential equation is

$y'(t)=te^{-t}$

It can be written as

$\frac{dy}{dt}=te^{-t}$

$dy=te^{-t}dt$

Integrate both sides.

$\int dy=\int te^{-t}dt$

Apply ILATE rule on right side. Here, t is first function and $e^{-t}$ is the second function.

$y=t\int e^{-t}-\int (\frac{d}{dt}t\int e^{-t})$

$y=-te^{-t}-\int (1\times (-e^{-t}))$         $\int e^{-x}=-e^{-x}+C$

$y=-te^{-t}+\int e^{-t}$

$y=-te^{-t}-e^{-t}+C$             .... (1)

Initial condition is y(1) = 1. It means at t=1 the value of y is 1.

$1=-(1)e^{-t}-e^{-(1)}+C$

$1=-e^{-1}-e^{-1}+C$

$1=-2e^{-1}+C$

$1=-\frac{2}{e}+C$

Add $\frac{2}{e}$ on both sides.

$1+\frac{2}{e}=C$

Substitute the value of C in equation (1).

$y=-te^{-t}-e^{-t}+1+\frac{2}{e}$

$y=-e^{-t}(t+1)+1+\frac{2}{e}$

Therefore, the solution of given initial value problem is $y=-e^{-t}(t+1)+1+\frac{2}{e}$.