Answer to Question #199735 in Differential Equations for priyanshu

Question

(3y−2xy^3)dx+(4x−3x^2y^2)dy=0

Answer 1

Given

$$(3y−2xy^3)dx+(4x−3x^2y^2)dy=0\\$$

then

$$M(x,y) = 3y−2xy^3\\ N(x,y) =4x−3x^2y^2\\ \frac{\partial M}{\partial y}=3-6xy^2\\ \frac{\partial N}{\partial x}=4-6xy^2\\ \frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}\\ \text{Showing that the equation is not exact.}\\$$

Let

$$x^ay^b$$

be the integrating factor such that the given D.E will be exact. Multiplying the given equation through with the I.F gives;

$$(3x^ay^{b+1}-2x^{a+1}y^{b+3})dx +(4x^{a+1}y^{b}-3x^{a+2}y^{b+2})dy\\$$

Since the equation is now exact, then

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

where:

$$\frac{\partial M}{\partial y} = 3(b+1)x^ay^{b}-2(b+3)x^{a+1}y^{b+2}\\ \frac{\partial N}{\partial x} = 4(a+1)x^ay^{b}-3(a+2)x^{a+1}y^{b+2}\\$$

By comparing the RHS of the equation, we get:

$3(b+1) = 4(a+1) \implies 4a-3b=-1 \qquad \cdots eqn(i)\\ -2(b+3)=-3(a+2) \implies b= \frac{3}{2}a \qquad \cdots eqn(ii)$

Solving the two equations simultaneously, we have:

$$a=2, b =3$$

Thus the IF for the equation to be exact is:

$$x^2y^3$$

Thus the exact form of the original equation is:

$$(3x^2y^4-2x^3y^5)dx+(4x^3y^3-3x^4y^5)dy=0$$

Integrating to get the solution to the equation:

$$\int (3x^2y^4-2x^3y^6)dx = x^3y^4-\frac{x^4y^6}{2}\\ \int (4x^3y^3-3x^4y^5)dy=x^3y^4-\frac{x^4y^6}{2}$$

Therefore, the general solution of the DE is:

$$x^3y^4-\frac{x^4y^6}{2}=c\\$$