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**Quiz-3 / Re: Q3 TUT 5101**

« **on:**October 12, 2018, 06:21:57 PM »

This is the hand written solution. I am still trying to figure out how to convert latex into plain text.

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This is the hand written solution. I am still trying to figure out how to convert latex into plain text.

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Since the equation is not exact, we need to find a function *µ*(x, y), such that

*µ*(3x^{2}y + 2xy + y^{3}) + *µ*(x^{2} + y^{2})y’ = 0 is exact.

M =*µ*(3x^{2}y + 2xy + y^{3}) N = *µ*(x^{2} + y^{2})

Then M_{y} = *µ*_{y}’(3x^{2}y + 2xy + y^{3}) + *µ*(3x^{2} + 2x + 3y^{2})

N_{x} = *µ*_{x}’(x^{2} + y^{2}) + *µ*(2x)

Let M_{y} = N_{x} , we get *µ*_{y}’(3x^{2}y + 2xy + y^{3}) + *µ*(3x^{2} + 3y^{2}) = *µ*_{x}’(x^{2} + y^{2})

First suppose*µ* is a function of x only.

Then*µ*_{y}’ = 0, we get

*µ*(3x^{2} + 3y^{2}) = *µ*_{x}’(x^{2} + y^{2})

3*µ* = ∂*µ*/∂x

By separable equation,

∫(1/*µ*) ∂*µ* = ∫3 ∂x

*ln*(µ) = 3x

*µ* = *e*^{3x} (the integration factor)

By theorem, there exist ϕ(x,y) , such that , ϕ_{x} = M, ϕ_{y} = N

ϕ = ∫*e*^{3x}(3x^{2}y + 2xy + y^{3}) = *e*^{3x}x^{2}y + (*e*^{3x}y^{3})/3 + h(y)

ϕ_{y} = *e*^{3x}x^{2} + h'(y) = *e*^{3x}x^{2} + *e*^{3x}y^{2}

Thus, h'(y) =*e*^{3x}y^{2}, h(y) = (*e*^{3x}y^{3})/3 + C

Therefore, the general solution is ϕ =*e*^{3x}x^{2}y + (*e*^{3x}y^{3})/3 = C

M =

Then M

N

Let M

First suppose

Then

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By separable equation,

∫(1/

By theorem, there exist ϕ(x,y) , such that , ϕ

ϕ = ∫

ϕ

Thus, h'(y) =

Therefore, the general solution is ϕ =

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Sorry, I posted to the wrong quiz question

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