- #1
flyingpig
- 2,579
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Homework Statement
i can solve them myself, but sometimes I am just lazy.How do you make Wolframalpha solve these exact, non-exact, other ODEs?
Ex.
[tex]2y^2 + 2y + 4x^2 + (2xy + x)y' = 0[/tex]
I tried using
DSolve[{(2*y[x]^2 + 2y[x] + 4x^2) + (2xy[x] + x)y'[x] = 0 }, y[x], x]
Not working?Problem
Solve [tex]2y^2 + 2y + 4x^2 + (2xy + x)y' = 0[/tex]
Work
[tex]\frac{\partial }{\partial y}(2y^2 + 2y + 4x^2) = 4y + 2[/tex]
[tex]\frac{\partial }{\partial x}(2xy + x) = 2y + 1[/tex]
[tex]\frac{4y+2-2y - 1}{2xy + x} = \frac{2y+1}{x(2y+1)} = \frac{1}{x}[/tex]
Then clearly the integrating factor is [tex]\mu (x) = x[/tex]
Now my question is, what is the solution that is lost? Is it y = -1/2? Because of [tex]\frac{2y+1}{x(2y+1)}[/tex]?
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