Question

\[\frac{\text dy}{\text dx}=\frac{2x+3y+4}{3x+y-1}\]

Answer 1

how can i check a solution?

Answer 2

ask wolf ...

Answer 3

change x = X+h and y=y+K ... so that 4 and -1 vanishes ... and it s a homogeneous DE

Answer 4

\[\frac{\text dy}{\text dx}=\frac{2x+3y+4}{3x+y-1}\]\[\text{let } y=Y+q,\qquad x=X+p,\qquad\frac{\text dy}{\text dx}=\frac{\text dY}{\text dX}\]\[\frac{\text dY}{\text dX}=\frac{2(X+p)+3(Y+q)+4}{3(X+p)+(Y+q)-1}\]\[\frac{\text dY}{\text dX}=\frac{2X+3Y+(2p+3q+4)}{3X+Y+(3p+q-1)}\]\[2p+3q+4=0\qquad3p+q-1=0\]\[\qquad\qquad\qquad\qquad\qquad\quad q=1-3p\]\[2p+3(1-3p)+4=0\qquad\qquad\qquad\qquad\]\[-7p+7=0\qquad\qquad\]\[p=1\qquad\qquad q=-2\]\[y+2=Y,\qquad x-1=X\]\[\frac{\text dY}{\text dX}=\frac{2X+3Y}{3X+Y}\]\[\frac{\text dY}{\text dX}=\frac{2+3\frac YX}{3+\frac YX}\]\[\text{let } \frac YX=V,\qquad Y= VX,\qquad\frac{\text dY}{\text dX}=V+X\frac{\text dV}{\text dX}\]\[V+X\frac{\text dV}{\text dX}=\frac{2+3V}{3+V}\]\[X\frac{\text dV}{\text dX}=\frac{2+3V}{3+V}-V\]\[X\frac{\text dV}{\text dX}=\frac{2+3V}{3+V}-V\frac{3+V}{3+V}\]\[X\frac{\text dV}{\text dX}=\frac{2+3V-3V-V^2}{3+V}\]\[X\frac{\text dV}{\text dX}=\frac{2-V^2}{3+V}\]\[\frac{3+V}{2-V^2}\text dV=\frac{\text dX}{X}\]\[\int\frac{3}{2-V^2}+\frac{V}{2-V^2}\text dV=\int\frac{\text dX}{X}\]\[-\frac {3}{2\times\sqrt 2}\ln\left|\frac{\sqrt 2+V}{\sqrt 2-V}\right|-\frac 12\ln\left|\sqrt {2}^2 -V^2\right|=\ln |X|+e^c\]\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2+\frac YX}{\sqrt 2-\frac YX}\right|+\ln\left|2-\left(\frac YX\right)^2\right|=-2\ln \left|cX\right|\]\[\frac {3}{\sqrt 2}\ln\left|\frac{X\sqrt 2+Y}{X\sqrt 2-Y}\right|+\ln\left|2\left(\frac {X}{X}\right)^2-\left(\frac YX\right)^2\right|=-2\ln \left|cX\right|\]\[\frac {3}{\sqrt 2}\ln\left|\frac{(x-1)\sqrt 2+(y+2)}{(x-1)\sqrt 2-(y+2)}\right|+\ln\left|\left(\frac {2(x-1)^2-(y+2)^2}{(x-1)^2}\right)^2\right|=-2\ln\left| c(x-1)\right|\]\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2 x+y+2-\sqrt 2}{\sqrt2x-y+2-\sqrt 2}\right|=\ln\left|\frac {(x-1)^2}{2x^2-4x+2-y^2-4y-4}\right|+\ln \left|\frac{1}{c(x-1)^2}\right|\]

Answer 5

\[\frac {3}{\sqrt 2}\ln\left|\frac{\sqrt 2 x+y+2-\sqrt 2}{\sqrt2x-y+2-\sqrt 2}\right|=-\ln\left|{c(2x^2-4x+2-y^2-4y-4)}\right|\]

Answer 6

you sure you did the integration rightfully?