## anonymous 5 years ago (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

1. anonymous

can u type the complete question pls?

2. anonymous

$(4x+3y^2)dx+(2xy)dy=0\rightarrow$ $M(x,y)=4x+3y^2, N(x,y)=2xy$ 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:$\mu(x,y)=x^my^n=x^2y^0=x^2\rightarrow m=2, n=0$