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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 $f(x,y)=C$

$\frac{dy}{dx}=\frac{5x^2}{4y}$

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

$0=0$

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

$\frac{-5x^{3}}{3}+g(y)$

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

$0+g'(y)$

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 $g'(y)$

$g'(y)=4y$

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

$\frac{-5x^{3}}{3}+2y^2=C_0$

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

$6y^2=C_0+5x^{3}$

Divide both sides of the equation by $6$

$y^2=\frac{C_0+5x^{3}}{6}$

Removing the variable's exponent

$y=\pm \sqrt{\frac{C_0+5x^{3}}{6}}$

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$y=\pm \frac{\sqrt{C_0+5x^{3}}}{\sqrt{6}}$

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\frac{\sqrt{C_0+5x^{3}}}{\sqrt{6}}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$y=\frac{\sqrt{C_0+5x^{3}}}{\sqrt{6}},\:y=\frac{-\sqrt{C_0+5x^{3}}}{\sqrt{6}}$
11

Find the explicit solution to the differential equation

$y=\frac{\sqrt{C_0+5x^{3}}}{\sqrt{6}},\:y=\frac{-\sqrt{C_0+5x^{3}}}{\sqrt{6}}$

Final Answer

$y=\frac{\sqrt{C_0+5x^{3}}}{\sqrt{6}},\:y=\frac{-\sqrt{C_0+5x^{3}}}{\sqrt{6}}$

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