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This is how I've resolve this ODE-1 :

$$(x^2 +y^2 +x) \, dx + xy \, dy=0$$

Check if the eq is exact: $${\partial M \over \partial y}={\partial \over \partial y}(x^2 +y^2 x)=2y$$ $${\partial N \over \partial x}={\partial \over \partial x}(xy)=y$$ $$A=2y \neq y=B$$

The equation is not exact. Trying to find the integrating factor : $$f=(A-B){1 \over N}$$ $$f=(2y-y){1 \over xy}={1 \over x}$$ Integrating factor: $$µ=e^{\int f(x) \, dx}=e^{\int {1 \over x} \, dx}=x$$ Now I can find the solution of the ODE by integrate the ODE with the integrating factor: $$\int x(x^2 +y^2 + x)\,dx + \int x(xy)\,dy=0$$ $$\int (x^3 +xy^2 +x^2)\,dx + \int x^2 y\,dy=0$$ $$\int x^3\,dx + \int xy^2 \,dx + \int x^2 \,dx + \int x^2 y\,dy=0$$ $${x^4 \over 4} + y^2 {x^2 \over 2} + {x^3 \over 3} + x^2 {y^2 \over 2} =0$$ $${x^4 \over 4} + {1 \over 2} x^2 y^2 + {x^3 \over 3} + {1 \over 2} x^2 y^2 =0$$ $$x^2 y^2 + {x^4 \over 4} + {x^3 \over 3} =0$$ The solution is : $12x^2 y^2 +3x^4 +4x^3 =C$

My teacher solution :

By integrating we get : ${x^4 \over 4}+{x^3 \over 3}+{1 \over 2}x^2 y^2=C$

The solution is : $3x^4 +4x^3 +6x^2 y^2 = C$

That's all she wrote...

Did I do something wrong or am I right? :s Thank you

Michael Hardy
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Marc-André Jean
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    you cannot integrate some therms over $x$ and others over $y$. This is your mistake. It is a line integral. – L.F. Cavenaghi Feb 06 '16 at 19:08
  • A line integral, are you sure? We didn't see that in my class yet I think – Marc-André Jean Feb 06 '16 at 21:39
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    once that now your euqation is exact the procedure to solve the equation is as follows: You have an equatin like: $Mdx + Ndy = 0.$ So, the condition exactness is: There is a function $f = f(x,y)$ such that $\partial_xf = M$. Then, $f(x,y) = \int Mdx + g(y)$. Ok, now derivating both sides, $\partial_yf(x,y) = \partial_y( \int Mdx ) + \partial_y g(y).$ But $\partial_yf(x,y) = N$. Then integrate $\partial_y g(y)$ to obtain $g(y)$ and then, put $f(x,y) = c$ once your equation is indeed, $df$. Now, once $df = 0$, it means, $f = c$, a constant function. – L.F. Cavenaghi Feb 06 '16 at 22:29
  • Ohh jezz yay true thnx! I think what I did at the end is the variable separable method, I crew up. – Marc-André Jean Feb 06 '16 at 22:38
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    There: if $U$ is my constant function $${\partial U \over \partial x}=x^3 +xy^2 +x^2$$ $${\partial U \over \partial y}=x^2y$$ $$U=\int x^2 y dy={1 \over 2}x^2 y^2 +h(x)$$ $${\partial \over \partial x}({1 \over 2}x^2 y^2 + h(x))$$ $$h'(x)=x^3 +xy^2 + x^2 - y^2x$$ $$h(x)= \int x^3 + xy^2 + x^2 - y^2$$ $$h(x)={1 \over 4}x^4 + {1/3}x^3$$ $$U={1 \over 2}x^2 y^2 + {1 \over 4}x^4 + {1 \over 3}x^3$$ $$U=6x^2y^2 +3x^4+4x^3$$

    I've skip some steps but I've now the right result :) thnx

     – Marc-André Jean Feb 07 '16 at 01:07
  • thats right! excelente! – L.F. Cavenaghi Feb 07 '16 at 15:47

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