Hi All,
Could not solve this one: 12-x-3y+(x-y+4)y'=0 , y(0)=1
I tried to find an integrating factor in order to make it an exact equation but did not succeed.
would appreciate your help!
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Hi All,
Could not solve this one: 12-x-3y+(x-y+4)y'=0 , y(0)=1
I tried to find an integrating factor in order to make it an exact equation but did not succeed.
would appreciate your help!
Put the differential equation into standard form. then Begin to find the integrating factor.
If I understand correctly the standard form has to be: y' + P(x)*y=Q(x) and specifically: y'-3y/(x-y+4)=(x-12)/(x-y+4)
but since there is a polynomial in the denominator of P(x) which includes y this is not the standard form.
Right?
I believe the standard form Cbarker was referring to is actually $\displaystyle \displaystyle \begin{align*} f(x,y)\,dx + g(x,y)\,dy = C \end{align*}$, which you can then find an integrating factor for to make the equation an exact equation.
