Question

How do you find the particular solution to $yy' - {e^x} = 0$ that satisfies y (0) = 4?

Answer 1

Hint: We will first use the method of separation of variables to find the solution to $yy' - {e^x} = 0$, then we will just put in x = 0 and y = 4 to get the value of constant and thus we have the required equation.

Complete step by step answer:
We are given that we are required to find the particular solution to $yy' - {e^x} = 0$ that satisfies y (0) = 4.
Given equation is $yy' - {e^x} = 0$.
We can write the given equation as follows:-
$ \Rightarrow y\dfrac{{dy}}{{dx}} - {e^x} = 0$
Taking ${e^x}$ from subtraction in the left hand side to addition in the right hand side, we will then obtain the following equation:-
$ \Rightarrow y\dfrac{{dy}}{{dx}} = {e^x}$
Taking dx from division in the left hand side to multiplication in the right hand side, we will then obtain the following equation:-
$ \Rightarrow ydy = {e^x}dx$
Integrating both the sides of the above equation, we will then obtain the following expression with us:-
\[ \Rightarrow \int {ydy} = \int {{e^x}dx} \]
Simplifying both the sides of the above mentioned expression, we will then obtain the following expression with us:-
\[ \Rightarrow \dfrac{{{y^2}}}{2} = {e^x} + c\]
Multiplying both the sides of the above equation by 2, we will then obtain the following equation with us:-
\[ \Rightarrow {y^2} = 2{e^x} + C\], where C = 2c ………………..(1)
Now, since, we are given that y (0) = 4 satisfies the equation, therefore, putting x = 0 and y = 4 in the above equation, we will then obtain the following equation with us:-
\[ \Rightarrow 16 = 2 + C\]
Thus, C = 14
Putting this in equation number 1, we will then obtain the following equation:-
\[ \Rightarrow {y^2} = 2{e^x} + 14\]
Hence, we have the required answer.

Note: The students must notice that the integration of exponential function ${e^x}$ is ${e^x}$ only without any change and the integration of algebraic function ${x^n}$ is given by $\dfrac{{{x^{n + 1}}}}{{n + 1}}$.
Therefore, after integration we got the equation as written.
The students must also notice that we used the method of variable separable because it was clearly visible to us that we can separate x and y’s from the equation so that we do not have any product term or division term related to them.