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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

Rewrite the differential equation using Leibniz notation

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

Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$

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

The differential equation $x^2e^y+\sin\left(x\right)-1dy1\left(y\cos\left(x\right)+2xe^y\right)dx=0$ 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 $f(x,y)=C$

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

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

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

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

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

Now take the partial derivative of $y\sin\left(x\right)+e^yx^2$ with respect to $y$ to get

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

Set $x^2e^y+\sin\left(x\right)-1$ and $\sin\left(x\right)+x^2e^y+g'(y)$ equal to each other and isolate $g'(y)$

$g'(y)=-1$
11

Find $g(y)$ integrating both sides

$g(y)=-y$
12

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

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

Then, the solution to the differential equation is

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

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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