Question

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

Answer 1

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.