Question

How do you solve the differential equation \[y' = {e^{ - y}}\left( {2x - 4} \right)\] , where \[y\left( 5 \right) = 0\] ?

Answer 1

Hint: The above differential equation is an example of a separable differential equation with an initial value. This equation can be factored into the product of two functions \[x\] and \[y\] , each of these depends upon only one variable. These types of equations can be written in the form \[y' = f\left( x \right)g\left( y \right)\] .

Formula used:
To solve this differential equation, we will use the formula \[\log {e^n} = n\log e\] . Also, keep in mind the value of \[\log e = 1\] .

Complete step by step solution:
The differential equation given to us is: \[y' = {e^{ - y}}\left( {2x - 4} \right)\] , where \[y\left( 5 \right) = 0\] .
We can write this as \[\dfrac{{dy}}{{dx}} = {e^{ - y}}\left( {2x - 4} \right)\] .
Now in the next step, we will multiple both the sides of the above equation by \[{e^y}\] , we get,
\[
  {e^y}\dfrac{{dy}}{{dx}} = {e^y}{e^{ - y}}\left( {2x - 4} \right) \\
   \Rightarrow {e^y}\dfrac{{dy}}{{dx}} = {e^{y - y}}\left( {2x - 4} \right) \\
   \Rightarrow {e^y}\dfrac{{dy}}{{dx}} = \left( {2x - 4} \right) \\
 \]
Now, we will again multiply both the sides by \[dx\] ,
\[
  dx{e^y}\dfrac{{dy}}{{dx}} = \left( {2x - 4} \right)dx \\
   \Rightarrow {e^y}dy = \left( {2x - 4} \right)dx \\
 \]
In the next step, we will integrate both the sides of the equation.
\[
  \int {} {e^y}dy = \int {} \left( {2x - 4} \right)dx \\
   \Rightarrow {e^y} = 2\dfrac{{{x^2}}}{2} - 4x + C \\
 \]
\[ \Rightarrow {e^y} = {x^2} - 4x + C\]
After integrating the equation, we will take natural log on both sides.
\[\log {e^y} = \log \left( {{x^2} - 4x + C} \right)\]
We know that, \[\log {e^n} = n\log e\] , so here we will replace \[n\] by \[y\] , and we get,
\[y\log e = \log \left( {{x^2} - 4x + C} \right)\]
We know, \[\log e = 1\] , so
\[y = \log \left( {{x^2} - 4x + C} \right)\]
Now, we want to find the value of C. we can find this value by using \[y\left( 5 \right) = 0\] .
\[y\left( 5 \right) = \log \left( {{{\left( 5 \right)}^2} - 4\left( 5 \right) + C} \right)\] (replacing \[x\] by \[5\] )
\[ = \log \left( {25 - 20 + C} \right)\]
\[ = \log \left( {5 + C} \right)\]
Now, \[y\left( 5 \right) = 0\] ,
\[\log \left( {5 + C} \right) = 0\]
\[
   \Rightarrow 5 + C = {e^0} \\
   \Rightarrow 5 + C = 1 \\
   \Rightarrow C = 1 - 5 \\
   \Rightarrow C = - 4 \\
 \]
Therefore, the value of \[C\] is \[ - 4\] .

Hence the final solution is \[y = \log \left( {{x^2} - 4x - 4} \right)\]

Note: To solve the above question of differential equation, we have applied the concept of separable differential equations. In order for a differential equation to be separable all the terms with \[x\] should be kept on one side of the equation and all the terms containing \[y\] , must be multiplied by the derivative. You must also remember the formulas of logarithm.