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### Topics - Junya Zhang

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##### Quiz-4 / Q4-T0701 / T0401
« on: March 02, 2018, 04:30:46 PM »
Verify that the given functions $y_1$ and $y_2$ satisfies the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$$t^2y'' - t(t+2)y' + (t+2)y = 2t^3, t>0; y_1(t)=t, y_2(t)=te^t$$

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##### Quiz-1 / Q1-T0701
« on: January 26, 2018, 02:02:06 PM »
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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