# Thread: dy/ dx + (y/x) = 2(x^2)

1. ## dy/ dx + (y/x) = 2(x^2)

I was told that the differential equation formed form this equation is EXACT. However , i found that the dM / dy not equal to dN/dx .... why ?

2. ## Re: dy/ dx + (y/x) = 2(x^2)

What about the other integrating factor?

3. ## Re: dy/ dx + (y/x) = 2(x^2)

Actually, don't you have your $M$ and $N$ transposed?

4. ## Re: dy/ dx + (y/x) = 2(x^2)

Originally Posted by Archie
Actually, don't you have your $M$ and $N$ transposed?
R(y) is (1/x)( -6x^2) = -6x? , so the integrating factor is e^-6xy ?

5. ## Re: dy/ dx + (y/x) = 2(x^2)

No. The equation should be $M \, \mathrm d x + N \, \mathrm d y = 0$, but you have $M \, \mathrm d y + N \, \mathrm d x = 0$.

6. ## Re: dy/ dx + (y/x) = 2(x^2)

Originally Posted by xl5899
I was told that the differential equation formed form this equation is EXACT. However , i found that the dM / dy not equal to dN/dx .... why ?
This equation is first order linear, just use an integrating factor...

\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{1}{x}\,y &= 2\,x^2 \end{align*}

Multiply both sides by the integrating factor \displaystyle \begin{align*} \mathrm{e}^{\int{\frac{1}{x}\,\mathrm{d}x}} = \mathrm{e}^{\ln{(x)}} = x \end{align*} and the equation becomes

\displaystyle \begin{align*} x\,\frac{\mathrm{d}y}{\mathrm{d}x} + y &= 2\,x^3 \\ \frac{\mathrm{d}}{\mathrm{d}x} \, \left( x\,y \right) &= 2\,x^3 \\ x\,y &= \int{ 2\,x^3\,\mathrm{d}x} \\ x\,y &= \frac{x^4}{2} + C \\ y &= \frac{x^3}{2} + \frac{C}{x} \end{align*}