Q46 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the non-homogeneous equation $(1-\mathrm{t})\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}-\mathrm{y}=(1-\mathrm{t}{)}^2$, given that  $\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}$is a solution to the corresponding homogeneous equation.

Step-by-Step Solution

Answer

The solution to the given non-homogeneous equation $(1-\mathrm{t})\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}-\mathrm{y}=(1-\mathrm{t}{)}^2$  is $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2\mathrm{t}$

.

Step 1: Find the value of Y

Given differential equation is $(1-\mathrm{t})\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}-\mathrm{y}=(1-\mathrm{t}{)}^2$then the standard form of the solution is $\mathrm{y}\text{'}\text{'}+\frac{\mathrm{t}}{1-\mathrm{t}}\mathrm{y}\text{'}-\frac{1}{1-\mathrm{t}}\mathrm{y}=\frac{1}{1-\mathrm{t}}$and $\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}$ is one of the solutions and $\mathrm{p}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{1-\mathrm{t}}$

Then ${\mathrm{y}}_2={\mathrm{y}}_1\left(\mathrm{t}\right)\int \frac{{\mathrm{e}}^{-\int \mathrm{p}\left(\mathrm{t}\right)}\mathrm{dt}}{{\mathrm{y}}_1^2\left(\mathrm{t}\right)}\mathrm{dt}$

Step 2: Simplification

Simplify the above.

$$\begin{gathered} \mathrm{t}\int \frac{{\mathrm{e}}^{-\int \frac{\mathrm{t}}{1-\mathrm{t}}}\mathrm{dt}}{{\left(\mathrm{t}\right)}^2}\mathrm{dt} \\ =\mathrm{t}\int \frac{{\mathrm{e}}^{\ln (\mathrm{t}-1)+\mathrm{t}}}{{\mathrm{t}}^2} \\ =\mathrm{t}\times \int \frac{(\mathrm{t}-1)\times {\mathrm{e}}^{\mathrm{t}}}{{\mathrm{t}}^2} \\ =\mathrm{t}\times \frac{{\mathrm{e}}^{\mathrm{t}}}{\mathrm{t}} \\ ={\mathrm{e}}^{\mathrm{t}} \end{gathered}$$

So, the solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2\mathrm{t}$