Author Topic: Q4-T0301/T0801  (Read 1537 times)

Victor Ivrii

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Q4-T0301/T0801
« on: March 02, 2018, 05:25:31 PM »
Find the general solution of the given differential equation.
$$y'' - 2y' + y = e^t/(1 + t^2).$$
« Last Edit: March 02, 2018, 05:39:34 PM by Victor Ivrii »

Meng Wu

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Re: Q4-T0301
« Reply #1 on: March 02, 2018, 05:35:25 PM »
$$y''-2y+y={e^t\over 1+t^2}$$
For homogeneous equation: $y''-2y+1=0$ $\\$
Characteristic equation: $$r^2-2r+1=0 \implies \cases{r_1=1\\r_2=1}$$
Complementary solution: $$\begin{align}y_c(t)&=c_1y_1(t)+c_2y_2(t)\\&=c_1e^t+c_2te^t\end{align}$$
For non-homogeneous equation: $y''-2y+y={e^t\over 1+t^2}$ $\\$
$p(t)=-2, q(t)=1,g(t)={e^t\over 1+t^2}$ are continuous everywhere.$\\$
Now, $$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t)\end{array}=\begin{array}{|c c|} e^t & te^t \\ e^t & (1+t)e^t \end{array}=e^{2t}\neq 0$$
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. $\\$
Therefore, $$\begin{align}u_1(t)=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}&=-\int{te^t\cdot {e^t\over1+t^2}\over e^2t}dt\\&=-\int{t\over1+t^2}dt \\&=-{1\over2}\ln(1+t^2)\end{align}$$
$$\begin{align}u_2(t)=\int{{y_1(t)g(t)\over W[y_1,y_2](t)}}&=\int{{e^t\cdot {e^t\over1+t^2}\over e^{2t}}}dt\\&=\int{1\over 1+t^2}dt\\&=\arctan t\end{align}$$
Hence, the particular solution is $y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$ $\\$
Thus, $$\begin{align}y_p(t)&=e^t\cdot (-{1\over2}\ln(1+t^2))+te^t\cdot \arctan t\\&=-{1\over2}e^t\ln(1+t^2)+te^t\arctan t\end{align}$$
Therefore, the general solution is:
$$\begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1e^t+c_2te^t-{1\over2}e^t\ln(1+t^2)+te^t\arctan t\end{align}$$