Find the integrating factor to make the differential equation exact, and then solve \[(3x^2y+2xy+y^3)\text d x+(x^2+y^2)\text dy=0\]

\[M=(3x^2y+2xy+y^3)\qquad\qquad N=(x^2+y^2)\]\[\frac{\partial M}{\partial y}=3x^2+2x+3y^2\qquad\qquad\frac{\partial N}{\partial x}=2x\]\[\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}\]\[\text{let}\quad R=R(x)\]\[\frac{\partial(R(x)M)}{\partial y}=\frac{\partial(R(x)N)}{\partial x}\]\[R(x)\frac{\partial M}{\partial y}=R(x)\frac{\partial N}{\partial x}+NR'(x)\]\[R(x)(3x^2+2x+3y^2)=R(x)(2x)+(x^2+y^2)R'(x)\]\[R(x)(3x^2+3y^2)=(x^2+y^2)R'(x)\]\[\frac{R'(x)}{R(x)}=\frac{3x^2+3y^2}{x^2+y^2}\]\[\frac{R'(x)}{R(x)}=3\]\[\int\frac{\text d R}{ R(x)}=\int3\text dx\]\[\ln R=3x\]\[R=e^{3x}\] \[MR(x)=e^{3x}(3x^2y+2xy+y^3)\qquad\qquad NR(x)=e^{3x}(x^2+y^2)\]\[\frac{\partial(R(x)M)}{\partial y}=e^{3x}\left(3x^2+2x+3y^2\right)\]\[\frac{\partial(R(x)N)}{\partial x}=3e^{3x}(x^2+y^2)+e^{3x}(2x)\]\[\frac{\partial(R(x)M)}{\partial y}=\frac{\partial(R(x)N)}{\partial x}\]\[f(x,y)=\int M\text dx+g(y)\]\[=\int e^{3x}(3x^2y+2xy+y^3)\text d x+g(y)\]

\[\cdots\]

you've definitely got the correct integrating factor

but then I would have used this to find f(x,y):\[f(x,y)=\int NR(x) dy \]which is much simpler

I assume you can solve from here?

\[f(x,y)=\int NR(x) dy=\int e^{3x}(x^2+y^2)dy=e^{3x}(x^2y+y^3/3)+h(x)\]

\[f(x,y)=\int NR(x)\text dy\]\[=\int e^{3x}(x^2+y^2)\text dy=e^{3x}\left(yx^2+\frac{y^3}3\right)+h(x)\]

h(x) can be found by doing a partial derivative of f(x,y) wrt to x and equating the result to MR(x)

use the fact that:\[MR(x)=\frac{\partial f(x,y)}{\partial x}\]

\[f(x,y)=e^{3x}\left(yx^2+\frac{y^3}3\right)+h(x)=e^{3x}(3x^2y+2xy+y^3)\]

no, you are are equating this:\[f(x,y)=MR(x)\]which is not correct. you need to differentiate f(x,y) with respect to x and then equate the result to MR(x), i.e.:\[\frac{\partial f(x,y)}{\partial x}=MR(x)\]

remember an exact differential equation can be written as:\[\frac{\partial f(x,y)}{\partial x}+\frac{\partial f(x,y)}{\partial y}\frac{dy}{dx}=0\]

i havent see it written in that form

after getting the integrating factor, your equation reduced to:\[MR(x)+NR(x)\frac{dy}{dx}=0\]

These online notes may be of help: http://www.cliffsnotes.com/study_guide/Exact-Equations.topicArticleId-19736,articleId-19710.html

These are also good: http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

\[f(x,y)=e^{3x}\left(yx^2+\frac{y^3}3\right)+h(x)\] \[\frac{\partial f}{\partial x}=3xe^{3x}(3x^2y+2xy+y^3)+e^{3x}(6xy+2y)+h'(x)\]\[=e^{3x}(9x^3y+6x^2y+3xy^3+6xy+2y)+h'(x)\] \[MR(x)=e^{3x}(3x^2y+2xy+y^3)\]

?,

your partial derivative does not look correct

ops

:)

\[f(x,y)=e^{3x}\left(yx^2+\frac{y^3}3\right)+h(x)\]\[\frac{\partial f}{\partial x}=3e^{3x}(x^2y+\frac {y^3}3)+e^{3x}(2xy)+h'(x)\]\[=e^{3x}(3x^2y+2xy+y^3)+h'(x)\] \[MR(x)=e^{3x}(3x^2y+2xy+y^3)\] \[h'(x)=0\]\[h(x)=c\]

\[f(x,y)=e^{3x}\left(yx^2+\frac{y^3}3\right)=-c=c_1\]

?is that the final solution

yes - looks fine to me.

Thank-you so Much

yw

for \(c=e^{\{-5,...5\}}\)

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