Question

How do you solve the differential equation $$y^{\prime }=e^{-y}\left(2x-4\right)$$ , 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 $$y^{\prime }=f\left(x\right)g\left(y\right)$$ .

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: $$y^{\prime }=e^{-y}\left(2x-4\right)$$ , where $$y\left(5\right)=0$$ .
We can write this as $${\displaystyle \frac{dy}{dx}}=e^{-y}\left(2x-4\right)$$ .
Now in the next step, we will multiple both the sides of the above equation by $$e^y$$ , we get,
$$\begin{gathered} e^y{\displaystyle \frac{dy}{dx}}=e^y{e}^{-y}\left(2x-4\right) \\ \Rightarrow e^y{\displaystyle \frac{dy}{dx}}=e^{y-y}\left(2x-4\right) \\ \Rightarrow e^y{\displaystyle \frac{dy}{dx}}=\left(2x-4\right) \end{gathered}$$
Now, we will again multiply both the sides by $$dx$$ ,
$$\begin{gathered} dx{e}^y{\displaystyle \frac{dy}{dx}}=\left(2x-4\right)dx \\ \Rightarrow e^{y}dy=\left(2x-4\right)dx \end{gathered}$$
In the next step, we will integrate both the sides of the equation.
$$\begin{gathered} \int e^{y}dy=\int \left(2x-4\right)dx \\ \Rightarrow e^y=2{\displaystyle \frac{x^2}{2}}-4x+C \end{gathered}$$
$$\Rightarrow e^y=x^2-4x+C$$
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, $$y\left(5\right)=0$$ ,
$$\log \left(5+C\right)=0$$
$$\begin{gathered} \Rightarrow 5+C=e^0 \\ \Rightarrow 5+C=1 \\ \Rightarrow C=1-5 \\ \Rightarrow C=-4 \end{gathered}$$
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.