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Exercise

$ycos\left(x\right)+2xe^y+\left(sinx+x^2e^y-1\right)y'=0$

Step-by-step Solution

1

Reescreva a equação diferencial usando a notação de Leibniz

$y\cos\left(x\right)+2xe^y+\frac{dy}{dx}\left(\sin\left(x\right)+x^2e^y-1\right)=0$
2

Rewrite the differential equation

$\frac{dy}{dx}\left(\sin\left(x\right)+x^2e^y-1\right)=-\left(y\cos\left(x\right)+2xe^y\right)$
3

Group terms with common factors

$\frac{dy}{dx}\left(x^2e^y+\sin\left(x\right)-1\right)=-\left(y\cos\left(x\right)+2xe^y\right)$
4

Rewrite the differential equation

$\frac{dy}{dx}=\frac{-\left(y\cos\left(x\right)+2xe^y\right)}{x^2e^y+\sin\left(x\right)-1}$
5

Reescreva a equação diferencial na forma padrão $M(x,y)dx+N(x,y)dy=0$

$x^2e^y+\sin\left(x\right)-1dy1\left(y\cos\left(x\right)+2xe^y\right)dx=0$
6

A equação diferencial $x^2e^y+\sin\left(x\right)-1dy1\left(y\cos\left(x\right)+2xe^y\right)dx=0$ é exata, pois está escrita em sua forma padrão $M(x,y)dx+N(x,y)dy=0$, onde $M(x,y)$ e $ N(x,y)$ constituem as derivadas parciais da função de duas variáveis ​​$f(x,y)$ e ambas satisfazem o teste de correção: $\displaystyle\frac{\partial M}{\partial y }=\frac{\partial N}{\partial x}$. Em outras palavras, suas segundas derivadas parciais são iguais. A solução geral da equação diferencial tem a forma: $f(x,y)=C$

$x^2e^y+\sin\left(x\right)-1dy1\left(y\cos\left(x\right)+2xe^y\right)dx=0$
7

Usando o teste de precisão, verificamos que a equação diferencial é exata

$\cos\left(x\right)+2xe^y=2e^yx+\cos\left(x\right)$
8

Integramos $M(x,y)$ em relação a $x$ para obter

$y\sin\left(x\right)+e^yx^2+g(y)$
9

Calcule a derivada parcial de $y\sin\left(x\right)+e^yx^2$ em relação a $y$ para obter

$\sin\left(x\right)+x^2e^y+g'(y)$
10

Igualamos $x^2e^y+\sin\left(x\right)-1$ e $\sin\left(x\right)+x^2e^y+g'(y)$ e então resolvemos para $g'(y)$

$g'(y)=-1$
11

Encontre $g(y)$ integrando ambos os lados

$g(y)=-y$
12

Encontramos nosso $f(x,y)$ e é equivalente a

$f(x,y)=y\sin\left(x\right)+e^yx^2-y$
13

Portanto, a solução da equação diferencial é

$y\sin\left(x\right)+e^yx^2-y=C_0$

Final answer to the exercise

$y\sin\left(x\right)+e^yx^2-y=C_0$

Try other ways to solve this exercise

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  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equations
  • Homogeneous Differential Equation
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
  • Integrate by parts
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