Answer to Question #199735 in Differential Equations for priyanshu

Question #199735

(3y−2xy^3)dx+(4x−3x^2y^2)dy=0


1
Expert's answer
2021-06-03T15:24:00-0400

Given


(3y2xy3)dx+(4x3x2y2)dy=0(3y−2xy^3)dx+(4x−3x^2y^2)dy=0\\

then


M(x,y)=3y2xy3N(x,y)=4x3x2y2My=36xy2Nx=46xy2MyNxShowing that the equation is not exact.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

xaybx^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;


(3xayb+12xa+1yb+3)dx+(4xa+1yb3xa+2yb+2)dy(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


My=Nx\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\\

where:


My=3(b+1)xayb2(b+3)xa+1yb+2Nx=4(a+1)xayb3(a+2)xa+1yb+2\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)    4a3b=1eqn(i)2(b+3)=3(a+2)    b=32aeqn(ii)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=3a=2, b =3

Thus the IF for the equation to be exact is:


x2y3x^2y^3

Thus the exact form of the original equation is:


(3x2y42x3y5)dx+(4x3y33x4y5)dy=0(3x^2y^4-2x^3y^5)dx+(4x^3y^3-3x^4y^5)dy=0

Integrating to get the solution to the equation:


(3x2y42x3y6)dx=x3y4x4y62(4x3y33x4y5)dy=x3y4x4y62\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:


x3y4x4y62=cx^3y^4-\frac{x^4y^6}{2}=c\\


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