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Ratling [72]
4 years ago
15

(4x+3y^2)dx+2xy*dy=0 by using integrating fector

Mathematics
1 answer:
Nataly_w [17]4 years ago
7 0
\underbrace{(4x+3y^2)}_M\,\mathrm dx+\underbrace{2xy}_N\,\mathrm dy=0

The ODE is exact if \dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}.

M_y=6y
N_x=2y

This is not the case, so look for an integrating factor \mu(x) such that

\dfrac\partial{\partial y}\mu M=\dfrac\partial{\partial x}\mu N

Since \mu is a function of x only, you have

\mu M_y=\mu'N+\mu N_x\implies\dfrac{\mu'}\mu=\dfrac{M_y-N_x}N\implies\mu=\exp\left(\displaystyle\int\frac{M_y-N_x}N\,\mathrm dx\right)

So, the integrating factor is

\mu=\exp\left(\displaystyle\int\frac{6y-2y}{2xy}\,\mathrm dx\right)=\exp\left(2\int\frac{\mathrm dx}x\right)=x^2

Now the ODE can be modified as

\underbrace{(4x^3+3x^2y^2)}_{M^*}\,\mathrm dx+\underbrace{2x^3y}_{N^*}\,\mathrm dy=0

Check for exactness:

{M^*}_y=6x^2y
{N^*}_x=6x^2y

so the modified ODE is indeed exact.

Now, you're looking for a solution of the form \Psi(x,y)=C, since differentiating via the chain rule yields

\dfrac{\mathrm d}{\mathrm dx}\Psi(x,y)=\Psi_x+\Psi_y\dfrac{\mathrm dy}{\mathrm dx}=0

Matching up components, you would have

\Psi_x=M^*=4x^3+3x^2y^2
\displaystyle\int\Psi_x\,\mathrm dx=\int(4x^3+3x^2y^2)\,\mathrm dx
\Psi=x^4+x^3y^2+f(y)

Differentiate this with respect to y to get

\Psi_y=2x^3y+f'(y)=2x^3y=N^*
f'(y)=0\implies f(y)=C_1

So the solution here is

\Psi(x,y)=x^4+x^3y^2+C_1=C\implies x^4+x^3y^2=C

Just for a final check, take the derivative to get back the original ODE:

\dfrac{\mathrm d}{\mathrm dx}[x^4+x^3y^2]=\dfrac{\mathrm d}{\mathrm dx}C
4x^3+3x^2y^2+2x^3y\dfrac{\mathrm dy}{\mathrm dx}=0
4x+3y^2+2xy\dfrac{\mathrm dy}{\mathrm dx}=0
(4x+3y^2)\,\mathrm dx+2xy\,\mathrm dy=0

so the solution is correct.
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