1
Solved example of first order differential equations
$\frac{dy}{dx}=\frac{5x^2}{4y}$
2
Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$
$4ydy-5x^2dx=0$
3
The differential equation $4ydy-5x^2dx=0$ is exact, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and they satisfy the test for exactness: $\displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form $f(x,y)=C$
$\frac{dy}{dx}=\frac{5x^2}{4y}$
Intermediate steps
Find the derivative of $M(x,y)$ with respect to $y$
$\frac{d}{dy}\left(-5x^2\right)$
The derivative of the constant function ($-5x^2$) is equal to zero
0
Find the derivative of $N(x,y)$ with respect to $x$
$\frac{d}{dx}\left(4y\right)$
The derivative of the constant function ($4y$) is equal to zero
0
4
Using the test for exactness, we check that the differential equation is exact
$0=0$
Intermediate steps
The integral of a function times a constant ($-5$) is equal to the constant times the integral of the function
$-5\int x^2dx$
Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$
$\frac{-5x^{3}}{3}$
Since $y$ is treated as a constant, we add a function of $y$ as constant of integration
$\frac{-5x^{3}}{3}+g(y)$
5
Integrate $M(x,y)$ with respect to $x$ to get
$\frac{-5x^{3}}{3}+g(y)$
Intermediate steps
The derivative of the constant function ($\frac{-5x^{3}}{3}$) is equal to zero
0
The derivative of $g(y)$ is $g'(y)$
$0+g'(y)$
6
Now take the partial derivative of $\frac{-5x^{3}}{3}$ with respect to $y$ to get
$0+g'(y)$
Intermediate steps
Simplify and isolate $g'(y)$
$4y=0+g$
$x+0=x$, where $x$ is any expression
$4y=g$
Rearrange the equation
$g=4y$
7
Set $4y$ and $0+g'(y)$ equal to each other and isolate $g'(y)$
$g'(y)=4y$
Intermediate steps
Integrate both sides with respect to $y$
$g=\int4ydy$
The integral of a function times a constant ($4$) is equal to the constant times the integral of the function
$g=4\int ydy$
Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$
$g=2y^2$
8
Find $g(y)$ integrating both sides
$g(y)=2y^2$
9
We have found our $f(x,y)$ and it equals
$f(x,y)=\frac{-5x^{3}}{3}+2y^2$
10
Then, the solution to the differential equation is
$\frac{-5x^{3}}{3}+2y^2=C_0$
Intermediate steps
Combine all terms into a single fraction with $3$ as common denominator
$\frac{-5x^{3}+2\cdot 3y^2}{3}=C_0$
Multiply $2$ times $3$
$\frac{-5x^{3}+6y^2}{3}=C_0$
Multiply both sides of the equation by $3$
$-5x^{3}+6y^2=3C_0$
We can rename $3C_0$ as other constant
$-5x^{3}+6y^2=C_0$
We need to isolate the dependent variable $y$, we can do that by subtracting $-5x^{3}$ from both sides of the equation
$6y^2=5x^{3}+C_0$
Divide both sides of the equation by $6$
$y^2=\frac{5x^{3}+C_0}{6}$
Removing the variable's exponent
$y=\pm \sqrt{\frac{5x^{3}+C_0}{6}}$
The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$
$y=\pm \frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}}$
As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign
$y=\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}},\:y=\frac{-\sqrt{5x^{3}+C_0}}{\sqrt{6}}$
11
Find the explicit solution to the differential equation
$y=\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}},\:y=\frac{-\sqrt{5x^{3}+C_0}}{\sqrt{6}}$
Final Answer
$y=\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}},\:y=\frac{-\sqrt{5x^{3}+C_0}}{\sqrt{6}}$