👉 Try now NerdPal! Our new math app on iOS and Android
  1. calculators
  2. Exact Differential Equation

Exact Differential Equation Calculator

Get detailed solutions to your math problems with our Exact Differential Equation step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

Go!
Symbolic mode
Text mode
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

1

Here, we show you a step-by-step solved example of exact differential equation. This solution was automatically generated by our smart calculator:

$x\:dx\:-\:y^2\:dy\:=\:0$
2

The differential equation $x\cdot dx-y^2dy=0$ 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 $f(x,y)=C$

$x\cdot dx-y^2dy=0$

Find the derivative of $M(x,y)$ with respect to $y$

$\frac{d}{dy}\left(x\right)$

The derivative of the constant function ($x$) is equal to zero

0

Find the derivative of $N(x,y)$ with respect to $x$

$\frac{d}{dx}\left(-y^2\right)$

The derivative of the constant function ($-y^2$) is equal to zero

0
3

Using the test for exactness, we check that the differential equation is exact

$0=0$

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 $n=1$

$\frac{1}{2}x^2$

Since $y$ is treated as a constant, we add a function of $y$ as constant of integration

$\frac{1}{2}x^2+g(y)$
4

Integrate $M(x,y)$ with respect to $x$ to get

$\frac{1}{2}x^2+g(y)$

The derivative of the constant function ($\frac{1}{2}x^2$) is equal to zero

0

The derivative of $g(y)$ is $g'(y)$

$0+g'(y)$
5

Now take the partial derivative of $\frac{1}{2}x^2$ with respect to $y$ to get

$0+g'(y)$

Simplify and isolate $g'(y)$

$-y^2=0+g$

$x+0=x$, where $x$ is any expression

$-y^2=g$

Rearrange the equation

$g=-y^2$
6

Set $-y^2$ and $0+g'(y)$ equal to each other and isolate $g'(y)$

$g'(y)=-y^2$

Integrate both sides with respect to $y$

$g=\int-y^2dy$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$g=-\int y^2dy$

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 $2$

$g=-\frac{y^{3}}{3}$

Multiplying the fraction by $-1$

$g=\frac{-y^{3}}{3}$
7

Find $g(y)$ integrating both sides

$g(y)=\frac{-y^{3}}{3}$
8

We have found our $f(x,y)$ and it equals

$f(x,y)=\frac{1}{2}x^2+\frac{-y^{3}}{3}$
9

Then, the solution to the differential equation is

$\frac{1}{2}x^2+\frac{-y^{3}}{3}=C_0$

Group the terms of the equation

$\frac{-y^{3}}{3}=C_0- \left(\frac{1}{2}\right)x^2$

Multiply the fraction and term in $- \left(\frac{1}{2}\right)x^2$

$\frac{-y^{3}}{3}=C_0-\frac{1}{2}x^2$

Multiplying the fraction by $x^2$

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

Multiply both sides of the equation by $3$

$-y^{3}=3\left(C_0+\frac{-x^2}{2}\right)$

Multiply both sides of the equation by $-1$

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

We can rename $-3\cdot C_0$ as other constant

$y^{3}=C_1+\frac{3x^2}{2}$

Removing the variable's exponent raising both sides of the equation to the power of $\frac{1}{3}$

$y=\sqrt[3]{C_1+\frac{3x^2}{2}}$
10

Find the explicit solution to the differential equation. We need to isolate the variable $y$

$y=\sqrt[3]{C_1+\frac{3x^2}{2}}$

Final answer to the problem

$y=\sqrt[3]{C_1+\frac{3x^2}{2}}$

Are you struggling with math?

Access detailed step by step solutions to thousands of problems, growing every day!