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Solve the equation
(35 x^(4) y+14 x^(5) y-2 y^(3)-4x y^(3) )dx + (7 x^(5) -6x y^(2) )dy leaving your answer in implicit form.

User Cybujan
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1 Answer

7 votes

\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0


M_y=35x^4+14x^5-6y^2-12xy^2

N_x=35x^4-6y^2


\frac{N_x-M_y}N=(-14x^5+12xy^2)/(7x^5-6xy^2)=-2

This suggests an integrating factor depending on
x only is possible, and given by


\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^(2x)

Distributing across the ODE, we end up with


\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^(2x)}_(M^*)\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^(2x)}_(N^*)\,\mathrm dy=0

The equation is now exact, with


{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^(2x)

Now we find the solution:


F_x=M^*

F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^(2x)\,\mathrm dx

F=(7x^5y-2xy^3)e^(2x)+f(y)


F_y=N^*

(7x^5-6xy^2)e^(2x)+f'(y)=(7x^5-6xy^2)e^(2x)

f'(y)=0

\implies f(y)=C

The general solution is then


F(x,y)=(7x^5y-2xy^3)e^(2x)=C
User POPI
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