MAT244-2018S > Quiz-1

Q1-T0701

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**Junya Zhang**:

Question: Find the solution of the given initial value problem.

$$y'-2y=e^{2t}, y(0)=2$$

Solution:

Notice that the given DE is a first order linear ODE.

Let $\mu(t)$ denote an integrating factor for the given DE.

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

Multiply the given DE by $\mu(t)$: note that $\mu(t) ≠ 0$ for all $t$

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

Simplify the equation:

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

Integrate both sides with respect to $t$:

$$e^{-2t}y = t + C$$

Isolate $y$:

$$y = (t+C)e^{2t}$$

Since $y(0)=2$, then $2 = (0+C)\cdot e^{0} = C\cdot 1 = C$

Thus, solution to the given IVP is

$$y = (t+2)e^{2t}$$

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