Question

2e^x is a particular integral for y'-y=2e^x. Is it true or false. Give reasons for your answer

Answer 1

Here we have, the differential equation $y'-y=2e^x$ where $y'=\frac{dy}{dx}$

Now, the co-efficient of $y$ is $-1$ .

So, let $\mu(x)=e^{\int-1dx}=e^{-x}$

Now, multiplying both sides of the differential equation by $\mu(x)$ we get:-

$$\mu(x)y'-\mu(x)y=2e^x\mu(x)\\ \Rightarrow e^{-x}\cdot\frac{dy}{dx}-ye^{-x}=2\\ \Rightarrow e^{-x}\cdot \frac{d}{dx}(y)+y\cdot\frac{d}{dx}(e^{-x})=2\\ \Rightarrow \frac{d}{dx}(ye^{-x})=2\\ \Rightarrow d(ye^{-x})=2dx\\$$

Integrating both sides we have:-

$$\int d(ye^{-x})=2\int dx+C \>\>\>\> (C(constant)\in\R)\\ \Rightarrow ye^{-x}=2x+C$$

Now, to check if $y=2e^x$ is a solution we substitute $y$ in place of the general solution:-

$2e^x\cdot e^{-x}=2x+C\\ \Rightarrow C=2-2x$

As $C$ is not a constant (i.e. independent of x) so $y=2e^x$ is not a solution for the given differential equation.