Question

(4x+3y^2)dx + (2xy)dy =0 by making use of the integrating factor m(x,y)= x^my^n where m,n are two real numbers to be found ? i tried to solve this question i found n=-2 m=-5 but it is wrong it does not satisfy the equation to be exact dif. equation

Answer 1

can u type the complete question pls?

Answer 2

\[(4x+3y^2)dx+(2xy)dy=0\rightarrow\]
\[M(x,y)=4x+3y^2, N(x,y)=2xy\]
Now check if they are exact:\[M_{y}=6y, N_{x}=2y\] They are not exact, you have to find an integrating factor:
\[\frac{d \mu}{dx}=\frac{M_{y}-N_{x}}{N(x,y)} \mu \rightarrow \frac {d \mu}{dx}=\frac{6y-2y}{2xy} \mu\]
\[\frac {d \mu}{dx}=\frac{6y-2y}{2xy} \mu=\frac {4y}{2xy} \mu= \frac{2}{x} \mu\]
\[\frac {d \mu}{\mu}=\frac{2dx}{x}\rightarrow \int\limits_{}\frac {d \mu}{\mu}=\int\limits_{}\frac{2dx}{x}\rightarrow \ln(\mu)=2\ln(x)\]
\[e^{\ln(\mu)}=e^{\ln(x^2)}\rightarrow \mu=x^2\]
This is your integrating factor, multiply your original equation with the integrating factor getting:
\[(4x^3+3x^2y^2)dx+(2x^3y)dy=0\]
\[M(x,y)=4x^3+3x^2y^2, N(x,y)=2x^3y\]
Check if they are exact once again:
\[M_{y}=6x^2y, N_{x}=6x^2y \rightarrow Exact\]
\[\psi(x,y)=\int\limits_{}2x^3ydy=x^3y^2+h(x)\]\[\psi_{x}=3x^2y^2+h'(x)\rightarrow M(x,y)=\psi_{x}\]
\[3x^2y^2+h'(x)=4x^3+3x^2y^2\rightarrow h'(x)=4x^3\]
\[\int\limits_{}h'(x)dx=\int\limits_{}4x^3dx \rightarrow h(x)=x^4+c\]
\[\psi(x,y)=x^3y^2+x^4\]
The solution is given implicitly by:
\[x^3y^2+x^4=C\]
Thus in your case, the integrating factor is:\[\mu(x,y)=x^my^n=x^2y^0=x^2\rightarrow m=2, n=0\]