Q35E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Given that $\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{,}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{t}$, and $\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}$ are solutions to the differential equation $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{y}\text{'}\mathbf{+}\mathbf{q}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$, find the solution to this equation that satisfies $\mathbf{y}\left(\mathbf{1}\right)\mathbf{=}\mathbf{2}$, $\mathbf{y}\text{'}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$ .

Step-by-Step Solution

Answer

The solution to the given equation $\mathrm{y}\text{'}\text{'}+\mathrm{p}\left(\mathrm{t}\right)\mathrm{y}\text{'}+\mathrm{q}\left(\mathrm{t}\right)\mathrm{y}=\mathrm{g}\left(\mathrm{t}\right)$ that satisfies $\mathrm{y}\left(1\right)=2$, $\mathrm{y}\text{'}\left(1\right)=0$ is $\mathrm{y}=1+2\mathrm{t}-{\mathrm{t}}^2$.

Step 1: Substitute the value for t

Given differential equation is $\mathrm{y}\text{'}\text{'}+\mathrm{p}\left(\mathrm{t}\right)\mathrm{y}\text{'}+\mathrm{q}\left(\mathrm{t}\right)\mathrm{y}=\mathrm{g}\left(\mathrm{t}\right)$ and solutions are $1+\mathrm{t},1+2\mathrm{t}$ and $1+3{\mathrm{t}}^2$.

So, the solution is $\mathrm{y}={\mathrm{c}}_1(1+\mathrm{t})+{\mathrm{c}}_2(1+2\mathrm{t})+{\mathrm{c}}_3\left(1+3{\mathrm{t}}^2\right)$

 

Now impose the initial conditions are:

$$\begin{gathered} \mathrm{y}\left(1\right)={\mathrm{c}}_1(1+1)+{\mathrm{c}}_2(1+2)+{\mathrm{c}}_3(1+3) \\ 2{\mathrm{c}}_1+3{\mathrm{c}}_2+4{\mathrm{c}}_3=2 ...\left(1\right) \end{gathered}$$ 

And 

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{c}}_1+2{\mathrm{c}}_2+6{\mathrm{c}}_3\mathrm{t} \\ \mathrm{y}\text{'}\left(1\right)={\mathrm{c}}_1+2{\mathrm{c}}_2+6{\mathrm{c}}_3 \\ {\mathrm{c}}_1+2{\mathrm{c}}_2+6{\mathrm{c}}_3=0 \end{gathered}$$

Step 2: Finding c 1 , c 2 , c 3

By the trial-and-error method, we will get ${\mathrm{c}}_1=\frac{2}{3},{\mathrm{c}}_2=\frac{2}{3},{\mathrm{c}}_3=-\frac{2}{3}$

So, the solution is $\begin{array}{rcl}\frac{2}{3}(1+\mathrm{t})+\frac{2}{3}(1+2\mathrm{t})-\frac{1}{3}\left(1+3{\mathrm{t}}^2\right)& =& \frac{2}{3}+\frac{2}{3}-\frac{1}{3}+\frac{2\mathrm{t}}{3}+\frac{4\mathrm{t}}{3}-{\mathrm{t}}^2\\ & =& 1+2\mathrm{t}-{\mathrm{t}}^2\end{array}$

 

Therefore, the solution is $\mathrm{y}=1+2\mathrm{t}-{\mathrm{t}}^2$.