Question

Determine whether the given differential equation is exact and solve it;

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{y}}\mathbf{=}\mathbf{-}\frac{\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{x}\mathbf{y}}{\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}}$

Short answer

We will solve the given differential equation

Step-by-step solution

Solve the given differential equation by substituting

Often the first step in solving the differential equation consists of transforming it into another differential equation by means of a substantial.

For example: Suppose we wish to transform the first order differential equation dy/dx=f(x,y) by the substitution y=g(x,u), where u is regarded as a function of the variable x.

Final Answer

Rewrite the equation as

$\begin{array}{c}\frac{dx}{dy}=-\frac{4y^2+6xy}{3y^2+2x}\\ \left(3y^2+2x\right)dx=-\left(4y^2+6xy\right)dy\\ \left(3y^2+2x\right)dx+\left(4y^2+6xy\right)dy=0\end{array}$

This is the form Mdx + Ndy = 0

M ( x, y ) =$3y^2+2x$ and N ( x, y) =$4y^2+6xy$

Now we have,

$\frac{\partial M}{\partial y}=6y=\frac{\partial N}{\partial x}$

which implies that the equation is exact, and so by Theorem 2.4.1 there exists a function f(x , y )such that

$\frac{\partial f}{\partial x}=3y^2+2x$and$\frac{\partial f}{\partial x}=4y^2+6xy$

Integrating the first part of this equation gives,

$\int \frac{\partial f}{\partial x}=\int \left(3y^2+2x\right)\Rightarrow f\left(x,y\right)=3x{y}^2+x^2+g\left(y\right)$

Now, we take the partial derivative this expression with respect to y and set the result equal to n(x,y) :

$\frac{\partial f}{\partial x}=6xy+g\text{'}\left(y\right)=4y^2+6xy:\#4257b2;\text{'}\text{'}>\leftarrow N(x,y)$

From the above equation, we get g’(y)=4y² . Integrating it gives g(y) = 4/3 y³

Therefore, the solution of the equation is

$f(x,y)=3x{y}^2+x^2+\frac{4}{3}{y}^3=c$

Hence, the final answer is:

$3x{y}^2+x^2+\frac{4}{3}{y}^3=c$