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Equazione differenziale esatta Calculator

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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

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

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