3 years ago

123°

Mathematics

1 answer:

zhuklara [117]3 years ago

8 0

Answer:

Supplementary 19

180-123=57

57÷3=19

Check:

123+57=180

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fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

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With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

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