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I missed a course and now I have to solve an ODE, I don't freaking know how :)

[tex](ax+by)dx+(kx+ly)dy=0[/tex]

So, I have to know on what condition this equation is exact (1), and I have to find the exact equation (2). Well if a, b, k and l = 0 then it's certainly over but i guess it's not really the point.

Let's say M(x,y) = ax+by and N(x,y) = kx+ly, IF i'm right, the ODE is exact when;

[tex]\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}[/tex]

Then;

[tex]\frac{\partial (ax+by)}{\partial y}=\frac{\partial (kx+ly)}{\partial x}[/tex]

[tex]b=k[/tex]

So my answer to the first question is; is it exact when b = k. And to show how sure I am (not...) I'll say c = b = k for the next part of this question. However, intuitively, i'd say the winning condition is when a and l = 0, and b = -k, (b=c,k=-c) then by integrating you'll got bxy+kxy = 0, so cxy-cxy=0, 0=0.

And then, I don't know what method to use to find the exact ODE.

[tex](ax+by)dx+(kx+ly)dy=0[/tex]

So, I have to know on what condition this equation is exact (1), and I have to find the exact equation (2). Well if a, b, k and l = 0 then it's certainly over but i guess it's not really the point.

Let's say M(x,y) = ax+by and N(x,y) = kx+ly, IF i'm right, the ODE is exact when;

[tex]\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}[/tex]

Then;

[tex]\frac{\partial (ax+by)}{\partial y}=\frac{\partial (kx+ly)}{\partial x}[/tex]

[tex]b=k[/tex]

So my answer to the first question is; is it exact when b = k. And to show how sure I am (not...) I'll say c = b = k for the next part of this question. However, intuitively, i'd say the winning condition is when a and l = 0, and b = -k, (b=c,k=-c) then by integrating you'll got bxy+kxy = 0, so cxy-cxy=0, 0=0.

And then, I don't know what method to use to find the exact ODE.

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