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kifflom [539]
3 years ago
12

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

exact find a function F(x,y) whose differential, dF(x,y) is the left hand side of the differential equation. That is, level curves F(x,y)=C, give implicit general solutions to the differential equation. If the equation is not exact, enter NE otherwise find F(x,y) (note you are not asked to enter C) F(x,y)= NE
Mathematics
1 answer:
zheka24 [161]3 years ago
4 0

Answer:

f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

Step-by-step explanation:

Let:

M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y

This is and exact equation, because:

\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}

So, define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)

The solution will be given by:

f(x,y)=C_1

Where C1 is an arbitrary constant

Integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y):

f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}

Substitute into \frac{\partial f(x,y)}{\partial y} =N(x,y)

2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y

Integrate \frac{dg(y)}{dy} with respect to y:

g(y)=\int\ {8y} \, dy =4y^2

Substitute g(y) into f(x,y):

f(x,y)=2x^2+4y^2+2xy

The solution is f(x,y)=C1

f(x,y)=2x^2+4y^2+2xy=C_1

Solving y using quadratic formula:

y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

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