How do you find the general solution of $4y{y}'-3{{e}^{x}}=0$ ?

Hint: We are given an equation which consists of a differential term, x-variable and y-variable. Such an equation is called a differential equation. We shall therefore separate the terms of the two variables on the two different sides of the equation and then integrate both sides accordingly to find the general solution.

Complete step-by-step solution:
Separable differential equations is the simplest form of differential equations. To solve such equations, we separate the terms of the x-variable with $dx$ and the terms of y-variable with $dy$ on the left-hand side and right-hand side of the equation. Then, we integrate both the sides of the equation individually and obtain the required equation.
We are given the equation $4y{y}'-3{{e}^{x}}=0$.
In order to separate the terms, we transpose the term $3{{e}^{x}}$ to the right hand side of the equation
$\Rightarrow 4y{y}'=3{{e}^{x}}$
${y}'$ is the first order derivative with respect to x and is also represented as $\dfrac{dy}{dx}$.
$\Rightarrow 4y\dfrac{dy}{dx}=3{{e}^{x}}$
Now, dividing both the sides by $dx$, we get
$\Rightarrow 4ydy=3{{e}^{x}}.dx$
Integrating both sides, we get
$\Rightarrow \int{4ydy}=\int{3{{e}^{x}}.dx}$
We know by the property of integration that $\int{{{x}^{n}}.dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$ and $\int{{{e}^{x}}.dx={{e}^{x}}+C}$ where $C$ is the constant of integration.
$\Rightarrow 4\dfrac{{{y}^{2}}}{2}+C=3{{e}^{x}}+C$
By combining both the constants, we get
$\Rightarrow 2{{y}^{2}}+3{{e}^{x}}+C=0$
Therefore, the general solution of $4y{y}'-3{{e}^{x}}=0$ is $2{{y}^{2}}+3{{e}^{x}}+C=0$.

Note: We have obtained the equation which consists of a constant term. This constant term raises ambiguity in the equation and thus it does not belong particularly to one curve only. However, if we put in the coordinates of any point lying on any particular curve in this equation, we can obtain the equation of that particular curve.