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The solution of $\dfrac{dy}{dx}={{e}^{3x+y}}$ given $y=0$ when $x=0$ is:
A. ${{e}^{3x}}+3{{e}^{-y}}=4$
B. ${{e}^{-y}}={{e}^{3x}}+4$
C. $3{{e}^{-y}}={{e}^{3x}}+12$
D. $y=\dfrac{{{e}^{3x+y}}}{3}$

Last updated date: 24th Jul 2024
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Hint: To obtain the solution of the differential equation we will use the Variable Separable method and integration of the exponent formula. Firstly we will separate the exponent in two terms with $x,y$ variable each then we will take each variable with its derivative and integrate them to get a solution. Finally we will put the given points to get the value of the constant and our desired answer.

Complete step-by-step solution:
The differential equation given to us is as follows:
We will use exponent property which is ${{e}^{a+b}}={{e}^{a}}\times {{e}^{b}}$ in above equation as follows:
$\dfrac{dy}{dx}={{e}^{3x}}\times {{e}^{y}}$
Now we will take each variable with its derivative as follows:
  & \dfrac{dy}{{{e}^{y}}}={{e}^{3x}}dx \\
 & \Rightarrow {{e}^{-y}}dy={{e}^{3x}}dx \\
Integrate both side of the above value as:
Now we will use exponent integration method which is $\int{{{e}^{ax}}dx}=\dfrac{{{e}^{ax}}}{a}$ above and get,
  & \dfrac{{{e}^{-y}}}{-1}=\dfrac{{{e}^{3x}}}{3}+c \\
 & \Rightarrow -{{e}^{-y}}=\dfrac{{{e}^{3x}}+3c}{3} \\
 & \Rightarrow -3{{e}^{-y}}={{e}^{3x}}+c \\
$\therefore 3{{e}^{-y}}+{{e}^{3x}}= -3c$….$\left( 1 \right)$
Next to find the value of constant put $y=0$ and $x=0$ in above value as follows:
  & 3{{e}^{0}}+{{e}^{3\times 0}}= -3c \\
 & \Rightarrow 3\times 1+1= -3c \\
 & \Rightarrow 3+1= -3c \\
 & \therefore c= \dfrac{-4}{3} \\
Putting the above value in equation (1) we get,
Hence correct option is (A).

Note: A differential equation is an equation that involves independent variable, dependent variable and the derivative of the dependent variables. We can solve a differential equation by many methods depending on the terms in it. Some methods are Variable separable method, Linear Integration, Homogeneous equation etc. Firstly we try to solve the differential equation using variable separable methods as it is easy. We just have to separate the variables and try to pair them up with their respective derivatives and integrate the whole equation and if it doesn’t work we try other methods.