Question

solve x2ydx-(x2+y3)dy =0​

Answer 1

Answer:

3y

3

x

3

=lny+C

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Step-by-step explanation:

x

2

ydx−(x

3

+y

3

)dy=0

dy

dx

=

x

2

y

x

3

+y

3

Putting x=vy

dy

dx

=v+y

dy

dv

dy

dx

=

x

2

y

x

3

+y

3

⇒v+y

dy

dv

=

v

2

y

3

v

3

y

3

+y

3

=

v

2

v

3

+1

y

dy

dv

=

v

2

v

3

+1

−v=

v

2

v

3

+1−v

3

=

v

2

1

⇒v

2

dv=

y

dy

(variable separable method)

Integrating both sides

∫v

2

dv=∫

y

dy

⇒

3

v

3

=lny+C

Putting v=

y

x

3y

3

x

3

=lny+C

Answer 2

EXPLANATION.

⇒ x²ydx - (x³ + y³)dy = 0.

As we know that,

We can write equation as,

⇒ x²ydx = (x³ + y³)dy.

⇒ dy/dx = (x²y)/(x³ + y³).

⇒ Substitute : y = vx.

⇒ dy/dx = v + x dv/dx.

$\implies \dfrac{dy}{dx} = \dfrac{x^{2} y}{(x^{3} + y^{3} )}$

$\implies v + x.\dfrac{dv}{dx} = \dfrac{x^{2} (vx)}{[x^{3} + (vx)^{3} ]}$

$\implies v + x.\dfrac{dv}{dx} = \dfrac{x^{3} v}{[x^{3} + v^{3}x^{3}] }$

$\implies v + x.\dfrac{dv}{dx} = \dfrac{x^{3} (v)}{x^{3}(1 + v^{3} ) }$

$\implies v + x.\dfrac{dv}{dx} = \dfrac{v}{1 + v^{3} }$

$\implies x\dfrac{dv}{dx} = \dfrac{v}{1 + v^{3} } - v$

$\implies x\dfrac{dv}{dx} = \dfrac{v - v(1 + v^{3}) }{1 + v^{3} }$

$\implies x\dfrac{dv}{dx} = \dfrac{v - v - v^{4} }{1 + v^{3} }$

$\implies x\dfrac{dv}{dx} = \dfrac{-v^{4} }{1 + v^{3} }$

$\implies \dfrac{1 + v^{3} }{v^{4} }dv = \dfrac{-dx}{x}$

Integrate both sides of the equation, we get.

$\implies \displaystyle \int \dfrac{1 + v^{3} }{v^{4} } dv \ = - \int \dfrac{dx}{x}$

$\implies \displaystyle \int \bigg( \dfrac{1}{v^{4}} + \dfrac{v^{3} }{v^{4} } \bigg) dv = - \int \dfrac{dx}{x}$

$\implies \displaystyle \int \dfrac{1}{v^{4} }dv \ + \int \dfrac{dv}{v} \ = - \int \dfrac{dx}{x}$

$\implies \displaystyle \int v^{-4} dv \ + \int \dfrac{dv}{v} \ = - \int \dfrac{dx}{x}$

$\implies \dfrac{v^{-4 + 1} }{-4 + 1} \ + ln|v| \ = - ln|x| + ln|C|$

$\implies \dfrac{v^{-3} }{-3} \ + ln|v| \ = - ln|x| + ln|C|$

$\implies \dfrac{-1}{3} . (v)^{-3} \ + ln|v| = - ln|x| + ln|C|$

$\implies \dfrac{-1}{3} .\bigg(\dfrac{y}{x} \bigg)^{-3} + ln|v| = - ln|x| + ln|C|$

$\implies -\dfrac{x^{3} }{3y^{3} } + ln \bigg|\dfrac{y}{x} \bigg| = - ln|x| + ln|C|$

$\implies -\dfrac{x^{3} }{3y^{3} } + ln|y| - ln|x| = - ln|x| + ln|C|$

$\implies -\dfrac{x^{3} }{3y^{3} } + ln|y| = ln|C|$

$\implies ln|y| + ln|C| = \dfrac{x^{3} }{3y^{3} }$

$\implies ln|yC| = \dfrac{x^{3} }{3y^{3} }$