👉 Prova ora NerdPal! La nostra nuova app di matematica su iOS e Android
  1. calcolatori
  2. Equazione Differenziale Esatta

Calcolatrice di Equazione differenziale esatta

Risolvete i vostri problemi di matematica con la nostra calcolatrice Equazione differenziale esatta passo-passo. Migliorate le vostre abilità matematiche con il nostro ampio elenco di problemi impegnativi. Trova tutte le nostre calcolatrici qui.

Go!
Modalità simbolica
Modalità testo
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:

$\left(4x+xy^2\right)dx\:+\left(y+x^2y\right)dy=0$
2

The differential equation $\left(4x+xy^2\right)dx+\left(y+x^2y\right)dy=0$ 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 $f(x,y)=C$

$\left(4x+xy^2\right)dx+\left(y+x^2y\right)dy=0$

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

$\frac{d}{dy}\left(4x+xy^2\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dy}\left(xy^2\right)$

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$x\frac{d}{dy}\left(y^2\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$2xy$

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

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

The derivative of a sum of two or more functions is the sum of the derivatives of each function

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$2yx$
3

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

$2xy=2yx$

Expand the integral $\int\left(4x+xy^2\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int4xdx+\int xy^2dx$

The integral of a function times a constant ($4$) is equal to the constant times the integral of the function

$4\int xdx+\int xy^2dx$

The integral of a function times a constant ($y^2$) is equal to the constant times the integral of the function

$4\int xdx+y^2\int xdx$

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 $n=1$

$4\cdot \left(\frac{1}{2}\right)x^2+y^2\int xdx$

Multiply the fraction and term in $4\cdot \left(\frac{1}{2}\right)x^2$

$2x^2+y^2\int xdx$

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 $n=1$

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

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

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

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

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

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dy}\left(\frac{1}{2}y^2x^2\right)$

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$\frac{1}{2}x^2\frac{d}{dy}\left(y^2\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

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

Multiply the fraction and term in $2\frac{1}{2}x^2y$

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

Any expression multiplied by $1$ is equal to itself

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

Divide $2$ by $2$

$x^2y$

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

$x^2y+g'(y)$
5

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

$x^2y+g'(y)$

Simplify and isolate $g'(y)$

$y+x^2y=x^2y+g$

Rearrange the equation

$x^2y+g=y+x^2y$

We need to isolate the dependent variable $g$, we can do that by simultaneously subtracting $x^2y$ from both sides of the equation

$g=y+x^2y-x^2y$

Cancel like terms $x^2y$ and $-x^2y$

$g=y$
6

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

$g'(y)=y$

Integrate both sides with respect to $y$

$g=\int ydy$

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 $n=1$

$g=\frac{1}{2}y^2$
7

Find $g(y)$ integrating both sides

$g(y)=\frac{1}{2}y^2$
8

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

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

Then, the solution to the differential equation is

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

Group the terms of the equation

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

Multiplying the fraction by $y^2x^2$

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

Multiplying the fraction by $y^2$

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

Combine fractions with common denominator $2$

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

Factor the polynomial $y^2x^2+y^2$ by it's greatest common factor (GCF): $y^2$

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

Multiply both sides of the equation by $2$

$y^2\left(x^2+1\right)=2\left(C_0-2x^2\right)$

Simplify $2\left(C_0-2x^2\right)$ using algebra

$y^2\left(x^2+1\right)=2\cdot -2x^2+C_1$

Multiply $2$ times $-2$

$y^2\left(x^2+1\right)=-4x^2+C_1$

Divide both sides of the equation by $x^2+1$

$y^2=\frac{-4x^2+C_1}{x^2+1}$

Removing the variable's exponent

$\sqrt{y^2}=\pm \sqrt{\frac{-4x^2+C_1}{x^2+1}}$

Cancel exponents $2$ and $1$

$y=\pm \sqrt{\frac{-4x^2+C_1}{x^2+1}}$

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{\frac{-4x^2+C_1}{x^2+1}}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$y=\sqrt{\frac{-4x^2+C_1}{x^2+1}},\:y=-\sqrt{\frac{-4x^2+C_1}{x^2+1}}$

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

$y=\frac{\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}},\:y=-\sqrt{\frac{-4x^2+C_1}{x^2+1}}$

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

$y=\frac{\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}},\:y=-\frac{\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}}$

Multiplying the fraction by $-1$

$y=\frac{\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}},\:y=\frac{-\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}}$

Combining all solutions, the $2$ solutions of the equation are

$y=\frac{\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}},\:y=\frac{-\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}}$
10

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

$y=\frac{\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}},\:y=\frac{-\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}}$

Risposta finale al problema

$y=\frac{\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}},\:y=\frac{-\sqrt{-4x^2+C_1}}{\sqrt{x^2+1}}$

Avete difficoltà in matematica?

Accedete a soluzioni dettagliate passo dopo passo per migliaia di problemi, che crescono ogni giorno!