Solved example of first order differential equations
Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$
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 $f(x,y)=C$
Find the derivative of $M(x,y)$ with respect to $y$
The derivative of the constant function ($-5x^2$) is equal to zero
Find the derivative of $N(x,y)$ with respect to $x$
The derivative of the constant function ($4y$) is equal to zero
Using the test for exactness, we check that the differential equation is exact
The integral of a function times a constant ($-5$) is equal to the constant times the integral of the function
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 $2$
Since $y$ is treated as a constant, we add a function of $y$ as constant of integration
Integrate $M(x,y)$ with respect to $x$ to get
The derivative of the constant function ($-\frac{5}{3}x^{3}$) is equal to zero
The derivative of $g(y)$ is $g'(y)$
Now take the partial derivative of $-\frac{5}{3}x^{3}$ with respect to $y$ to get
Simplify and isolate $g'(y)$
$x+0=x$, where $x$ is any expression
Rearrange the equation
Set $4y$ and $0+g'(y)$ equal to each other and isolate $g'(y)$
Integrate both sides with respect to $y$
The integral of a function times a constant ($4$) is equal to the constant times the integral of the function
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 $n=1$
Find $g(y)$ integrating both sides
We have found our $f(x,y)$ and it equals
Then, the solution to the differential equation is
We need to isolate the dependent variable $y$, we can do that by simultaneously subtracting $-\frac{5}{3}x^{3}$ from both sides of the equation
Divide both sides of the equation by $2$
Simplifying the quotients
Removing the variable's exponent
Cancel exponents $2$ and $\frac{1}{2}$
As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{\frac{\frac{5}{3}x^{3}+C_0}{2}}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign
Combining all solutions, the $2$ solutions of the equation are
Find the explicit solution to the differential equation. We need to isolate the variable $y$
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