Question

Let $y=y\left( x \right)$ be the solution of the differential equation $\dfrac{dy}{dx}+2y=f\left( x \right)$, where $f\left( x \right)=\left\{ \begin{matrix}
   1,\text{ }x\in \left( 0,1 \right) \\
   0,\text{ otherwise} \\
\end{matrix} \right.$
If $y\left( 0 \right)=0$ then $y\left( \dfrac{3}{2} \right)$ is

Answer 1

Hint: We solve this question by first comparing the given differential equation with $\dfrac{dy}{dx}+yP\left( x \right)=Q\left( x \right)$. Then we solve it by multiplying the equation with integrating factor by finding the integrating factor using the formula $I.F={{e}^{\int{P\left( x \right)dx}}}$. Then we solve the differential equation and find the values of the function $y=y\left( x \right)$ when $x\in \left( 0,1 \right)$ and when $x\notin \left( 0,1 \right)$. Then we substitute the value $x=0$ in $y\left( x \right)$ to find the value of function. Then we substitute the value $x=\dfrac{3}{2}$ in $y\left( x \right)$ to find the value $y\left( \dfrac{3}{2} \right)$.

Complete step-by-step answer:
Let us consider the given differential equation $\dfrac{dy}{dx}+2y=f\left( x \right)$.
As we see, it is in the form of $\dfrac{dy}{dx}+yP\left( x \right)=Q\left( x \right)$.
If any differential equation is of the above form then we multiply it with an integrating factor which is given by
$I.F={{e}^{\int{P\left( x \right)dx}}}$
Then we solve the differential equation as
$y\times \left( I.F \right)=y{{e}^{\int{P\left( x \right)dx}}}=\int{{{e}^{\int{P\left( x \right)dx}}}Q\left( x \right)}dx=\int{\left( I.F \right)\times Q\left( x \right)}dx$
Here, by comparing the given differential equation with the above form of differential equation we can say that,
$P\left( x \right)=2$ and $Q\left( x \right)=f\left( x \right)$
So, now let us find the integrating factor for our given differential equation,
$\begin{align}
  & \Rightarrow I.F={{e}^{\int{2dx}}} \\
 & \Rightarrow I.F={{e}^{2\int{dx}}} \\
 & \Rightarrow I.F={{e}^{2x}} \\
\end{align}$
So, we get the solution of the differential equation as
$y{{e}^{2x}}=\int{{{e}^{2x}}f\left( x \right)dx}$
As we are given that $f\left( x \right)=\left\{ \begin{matrix}
   1,\text{ }x\in \left( 0,1 \right) \\
   0,\text{ otherwise} \\
\end{matrix} \right.$, let us substitute the value of $f\left( x \right)$ in the above solution of differential equation.
When $x\in \left( 0,1 \right)$
$\begin{align}
  & \Rightarrow y{{e}^{2x}}=\int{{{e}^{2x}}\times 1dx} \\
 & \Rightarrow y{{e}^{2x}}=\int{{{e}^{2x}}dx} \\
\end{align}$
Now let us consider the formula $\int{{{e}^{ax}}dx}=\dfrac{{{e}^{ax}}}{a}+c$. Using this we can write the above equation as,
$\Rightarrow y{{e}^{2x}}=\dfrac{{{e}^{2x}}}{2}+c$
Now let us divide the above equation with ${{e}^{2x}}$.
$\Rightarrow y=\dfrac{1}{2}+c{{e}^{-2x}}$
As, we are given that $y=y\left( x \right)$ we can substitute it in the above equation.
$\Rightarrow y\left( x \right)=\dfrac{1}{2}+c{{e}^{-2x}}$
So, we get the value of $y\left( x \right)$ in the interval $x\in \left( 0,1 \right)$ as $y\left( x \right)=\dfrac{1}{2}+c{{e}^{-2x}}$.
When $x\notin \left( 0,1 \right)$
$\begin{align}
  & \Rightarrow y{{e}^{2x}}=\int{{{e}^{2x}}\times 0dx} \\
 & \Rightarrow y{{e}^{2x}}=\int{0dx} \\
\end{align}$
Now let us consider the formula $\int{0dx}=\text{constant}$. Using this we can write the above equation as,
$\begin{align}
  & \Rightarrow y{{e}^{2x}}=C \\
 & \Rightarrow y=C{{e}^{-2x}} \\
\end{align}$
So, we get the function $y\left( x \right)$ as
$y\left( x \right)=\left\{ \begin{matrix}
   \dfrac{1}{2}-c{{e}^{-2x}},\text{ }x\in \left( 0,1 \right) \\
   C{{e}^{-2x}},\text{ otherwise} \\
\end{matrix} \right.$
We are given that $y\left( 0 \right)=0$.
When $x=0$ the function is $y\left( x \right)=C{{e}^{-2x}}$. So, substituting the value $x=0$, we get
$\begin{align}
  & \Rightarrow y\left( 0 \right)=0=C{{e}^{-2\left( 0 \right)}} \\
 & \Rightarrow C{{e}^{0}}=0 \\
 & \Rightarrow C=0 \\
\end{align}$
Then the function becomes $y\left( x \right)=\left\{ \begin{matrix}
   \dfrac{1}{2}-c{{e}^{-2x}},\text{ }x\in \left( 0,1 \right) \\
   0,\text{ otherwise} \\
\end{matrix} \right.$
As we are asked to find the value of $y\left( \dfrac{3}{2} \right)$, let us substitute the value $x=\dfrac{3}{2}$ in the above function $y\left( x \right)$.
As $\dfrac{3}{2}\notin \left( 0,1 \right)$, value of $y\left( \dfrac{3}{2} \right)$ is 0.
Hence, we get the value of $y\left( \dfrac{3}{2} \right)$ as 0.
Hence, the answer is 0.

Note: The main mistake one does is one might not check that when we are given x=0, it does not belong to the interval (0,1) and take the value of function as $y\left( x \right)=\dfrac{1}{2}+c{{e}^{-2x}}$ and find the value of c and then find the value of $y\left( \dfrac{3}{2} \right)$. But it is wrong. Here we need to consider the function when x does not belong to (0,1).