Need Help?

Subscribe to Differential Equation

  • Notes
  • Comments & Questions

$$\left(y^{2}+y x\right) d x-x^{2} d y=0$$

$$\left(M(x, y) d x_{+} N(x, y) d y=0\right.$$ homog

$$u=f(x) \quad y=u x$$

$$d y=x d u+u d x$$

$$\left(u^{2} x^{2}+u x^{2}\right) d x-x^{2}(x d u+u d x)=0$$

$$u^{2} x^{2} d x+u x^{2} d x-x^{3} d u-x^{2} u d x=0$$

$$u^{2} x^{2} d x-x^{3} d u=0$$

$$\div u^{2} x^{3}$$

$$\frac{u^{2} x^{2}}{u^{2} x^{3}} d x-\frac{x^{3}}{u^{2} x^{3}} d u=0$$

$$\frac{d x}{x}-\frac{d u}{u^{2}}=0$$

$$\frac{d x}{x}=\frac{d u}{u^{2}}$$

$$\int \frac{d x}{x}=\ln |x|$$

$$\int u^{-2} d u=-u^{-1}$$

$$\ln |x|=-u^{-1}+C$$

$$y=u x \quad u=\frac{y}{x} \quad u^{-1}=\frac{x}{y}$$

$$\ln |x|=-\frac{x}{y}+c$$

$$y \ln |x|+x=c y$$

$$\left(y^{2}+y x\right) d x-x^{2} d y=0$$

$$V=f(y) \quad x=v y \quad d x=v d y+y d v$$

$$\left(y^{2}+v y^{2}\right)(v d y+y d v)-v^{2} y^{2} d y=0$$

$$y^{2} v d y+y^{3} d v+v^{2} y^{2} d y+v y^{3} d v_{-} v^{2} y^{2} d y = 0$$

$$y^{2} v d y+\left(y^{3}+v y^{3}\right) d v=0$$

$$\div V y^{3}$$

$$\frac{y^{2} v}{v y^{3}} d y+\left(\frac{y^{3}}{v y^{3}}+\frac{v y^{3}}{v y^{3}}\right) d v=0$$

$$\frac{d y}{y}+\left(\frac{1}{v}+1\right) d v=0$$

$$\frac{d y}{y}=-\left(\frac{1}{v}+1\right) d v$$

$$\ln |y|=-\ln |v|-v+c$$


$$\ln |y|=-\ln |x / y|-x / y+c$$

$$=-[\ln |x|-\ln |y|]-x / y+c$$

$$\ln |y|=-\ln |x|+\ln |y|-x / y+c$$

$$\ln |x|+x / y=c$$

$$y \ln |x|+x=c y$$

$$y d x+x(\ln x-\ln y-1) d y=0 \quad y(1)=e$$

$$v=f(y) \quad x=v y \quad d x=v d y+y d v$$

$$y(v d y+y d v)+v y(\ln (v y)-\ln y-1) d y=0$$

$$\ln v y=\ln v+\ln y$$

$$y v d y+y^{2} d v+v y(\ln v+\ln y-\ln y-1) d y=0$$

$$y v d y+y^{2} d v+v y \ln v d y-v y d y=0$$

$$y^{2} d v+v y \ln v d y=0$$

$$\div y^{2} v \ln v$$

$$\frac{y^{2}}{y^{2} v \ln v} d v+\frac{v y \ln v}{y^{2} v \ln v} d y=0$$

$$\frac{d v}{v \ln v}+\frac{d y}{y}=0$$

$$\frac{d v}{v \ln v}=-\frac{d y}{y}$$

$$\int \frac{(1 / v) dv}{\ln v}=\ln (\ln v)$$

$$\ln |\ln | v| |=-\ln |y|+c$$

$$\ln |\ln | x / y| |=-\ln |y|+c$$

$$e^{{\ln |\ln | x / y} | |}=e^{-\ln |y|+c}$$

$$\ln \left|x \right/ y |=e^{c-\ln |y|}=\frac{e^{c}}{e^{\ln |y|}}$$

$$\ln \left|x / y\right|=\frac{e^{c}}{y}$$

$$y \ln \left|x / y\right|=C_{1}$$


$$C_{1}=e \ln \left|\frac{1}{e}\right|$$


$$y \ln |x/y |=-e$$

No comments yet

Join the conversation

Join Notatee Today!