Author Topic: Q3-T0501  (Read 710 times)

Victor Ivrii

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Q3-T0501
« on: February 10, 2018, 05:18:09 PM »
Find the general solution of the given differential equation.
$$
2y'' - 3y' + y = 0.
$$

Junjie Zhang

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Re: Q3-T0501
« Reply #1 on: February 10, 2018, 05:57:47 PM »
Substitution of the assumed solution $y=e^{rt}$ results in the characteristic equation $$2r^2-3r+1=0$$
 The roots of the equation are $r = \frac{1}{2}, 1$. Hence the general solution is $y = c_{1}e^{\frac{t}{2}}+c_{2}e^{\frac{3t}{2}}$
« Last Edit: February 10, 2018, 06:00:49 PM by Junjie Zhang »

Meng Wu

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Re: Q3-T0501
« Reply #2 on: February 11, 2018, 09:29:22 AM »
$$2y’’-3y’+y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$2r^2-3r+1=(2r-1)(r-1)=0$$
Hence,
$$\cases{r_1={1\over2}\\r_2=1}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{{1\over2}t}+c_2e^{t}$$