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Solve the homogeneous ODE (or bring it to homogeneous ODE and solve) $$y'=\frac{1-3x-3y}{1+x+y}$$

How should I approach this?

2 Answers 2

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$$y'=\frac { 1-3x-3y }{ 1+x+y } \\ \\ \\ y=z-x\\ { y }^{ \prime }={ z }^{ \prime }-1\\ { z }^{ \prime }-1=\frac { 1-3z }{ 1+z } \\ { z }^{ \prime }=\frac { 1-3z }{ 1+z } +1=\frac { 2-2z }{ 1+z } \\ \\ \int { \frac { 1+z }{ 1-z } dz } =2\int { dx } \\ \int { \left( 1+\frac { 2 }{ z-1 } \right) dz } =-2x+C\\ z+2\ln { \left| z-1 \right| = -2x+C } \\ x+y+2\ln { \left| x+y-1 \right| =-2x+C } \\ $$

$$2\ln { \left| x+y-1 \right| =-3x-y+C } \\ \\ $$

  • 0
    I have came to $2ln|1-z|+z=-2x+c$ how can I isolate $z$? – 2017-01-30
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    I will put full answer here to be more clear – 2017-01-30
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    How did you get to $ \int { \left( 1+\frac { 2 }{ z-1 } \right) dz } =-2x+C$ – 2017-01-30
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    @gbox $\int { \frac { 1+z }{ 1-z } dz } =-\int { \frac { z+1 }{ z-1 } } dz=-\int { \left( \frac { z-1+2 }{ z-1 } \right) dz=-\int { \left( 1+\frac { 2 }{ z-1 } \right) dz } =2x+C } \\ \int { \left( 1+\frac { 2 }{ z-1 } \right) dz } =-2x+C$ – 2017-01-30
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    What a performance in the category "proof without words"...:) – 2017-01-30
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    @JeanMarie,thank you) – 2017-01-30
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Let $u=1+x+y$, $\frac{du}{dx}=1+\frac{dy}{dx}$. So your equation becomes $$\frac{du}{dx}-1=\frac{4}{u}-3$$ $$\frac{du}{dx}=\frac{4}{u}-2=\frac{4-2u}{u}$$ $$\frac{u}{4-2u}\frac{du}{dx}=1$$ $$\int\frac{u}{4-2u}\frac{du}{dx}\,dx=x+C$$ $$\int\frac{u}{4-2u}\,du=x+C$$

Now compute the integral, isolate $u$, back substitute to bring back $y$, and isolate $y$.