Question

Solve the initial value problem:
$dy={{e}^{2x+y}}dx,y\left( 0 \right)=0$

Answer 1

Hint: Here, we will first take the function involving x on one side and the function involving y to the other side. After obtaining the solution, we can find the value of the constant of integration using the given condition.

Complete step-by-step answer:
We know that an initial value problem is an ordinary differential equation together with an initial condition which specifies the value of the unknown function at a given point in the domain.
Here, the differential equation given to us is:
$dy={{e}^{2x+y}}dx$
We can also write this equation as:
$\dfrac{dy}{dx}={{e}^{2x}}.{{e}^{y}}$
Now, to make the similar terms on same side, we can also write this as:
$\dfrac{dy}{{{e}^{y}}}={{e}^{2x}}.dx$
On integrating both sides, we get:
$\int_{{}}^{{}}{{{e}^{-y}}dy=\int_{{}}^{{}}{{{e}^{2x}}dx..........\left( 1 \right)}}$
Since, we know that the integration of ${{e}^{x}}$ is given as $\int_{{}}^{{}}{{{e}^{x}}dx={{e}^{x}}+c}$
Now, let us consider that $t=-y$.
On differentiating w.r.t y, we get:
$dy=-dt$
So,
$\int_{{}}^{{}}{{{e}^{-y}}dy=\int_{{}}^{{}}{{{e}^{t}}.\left( -dt \right)=-\int_{{}}^{{}}{{{e}^{t}}dt}=-{{e}^{t}}+c=-{{e}^{-y}}+{{c}_{1}}}}$
Similarly let us consider that $k=2x$ .
On differentiating w.r.t. x, we get:
$\begin{align}
  & \Rightarrow 2=\dfrac{dk}{dx} \\
 & \Rightarrow 2dx=dk \\
\end{align}$
So,
$\int_{{}}^{{}}{{{e}^{2x}}dx=\int_{{}}^{{}}{{{e}^{k}}.\dfrac{dk}{2}=\dfrac{1}{2}\int_{{}}^{{}}{{{e}^{k}}dk=\dfrac{1}{2}{{e}^{k}}+{{c}_{2}}=\dfrac{1}{2}{{e}^{2x}}+{{c}_{2}}}}}$
On putting the corresponding values in equation (1), we get:
$\begin{align}
  & -{{e}^{-y}}+{{c}_{1}}=\dfrac{{{e}^{2x}}}{2}+{{c}_{2}} \\
 & \Rightarrow \dfrac{{{e}^{2x}}}{2}+{{e}^{-y}}={{c}_{1}}-{{c}_{2}}=c...........\left( 2 \right) \\
\end{align}$
Here, c is only another integration constant and it is equal to ${{c}_{1}}-{{c}_{2}}$.
Since, it is given that $y\left( 0 \right)=0$.
So, on substituting x = 0 and y = 0 in equation (2), we get:
$\begin{align}
  & \dfrac{{{e}^{0}}}{2}+{{e}^{0}}=c \\
 & \Rightarrow c=\dfrac{1}{2}+1=\dfrac{3}{2} \\
\end{align}$
Hence, the solution of the given differential equation is $\dfrac{{{e}^{2x}}}{2}+{{e}^{-y}}=\dfrac{3}{2}$.

Note: Students should note here that we have written ${{c}_{1}}-{{c}_{2}}$ as another constant C because the difference of two constants will also give us a constant. Students should remember the formulas of integration.