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Greeley [361]
3 years ago
7

I was just wondering this is one of my questions in my math books and i needed help

Mathematics
1 answer:
just olya [345]3 years ago
7 0

Answer:

B is true

Step-by-step explanation:

5 is a coefficient of x^0

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\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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8+\ln9=C

so

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

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Step-by-step explanation:

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