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exis [7]
3 years ago
6

Solve the equation

3%7D%20%29dx%20%2B%20%287%20%20x%5E%7B5%7D%20-6x%20y%5E%7B2%7D%20%29dy" id="TexFormula1" title="(35 x^{4} y+14 x^{5} y-2 y^{3}-4x y^{3} )dx + (7 x^{5} -6x y^{2} )dy" alt="(35 x^{4} y+14 x^{5} y-2 y^{3}-4x y^{3} )dx + (7 x^{5} -6x y^{2} )dy" align="absmiddle" class="latex-formula"> leaving your answer in implicit form.
Mathematics
1 answer:
jeka943 years ago
7 0
\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

\dfrac{N_x-M_y}N=\dfrac{-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
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