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My book says:

"Differential equations of the form $$\frac{\mathrm{d} y}{\mathrm{d} x}=f(x,y)$$, where $$f(x,y)$$ is homogenous function (function is homogenous if $$f(tx,ty)=t^k f(x,y)$$) can be written in form $$\frac{\mathrm{d} y}{\mathrm{d} x}=F(\frac{y}{x})$$ and transformed to differential equation with separate variables using substitution $$\frac{y}{x}=z(x)$$."

Here is my differential equation and how I tried to solve it:

$$(x+y)^2 \frac{\mathrm{d} y}{\mathrm{d} x}=r^2$$

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}$$

$$f(x,y)=\frac{r^2}{(x+y)^2}$$

$$\rightarrow f(tx,ty)=\frac{r^2}{(tx+ty)^2}=\frac{r^2}{(t(x+y))^2}=\frac{r^2}{t^2(x+y)^2}=t^{-2}\frac{r^2}{(x+y)^2}$$ so our function is homogenous and we can solve our diff. eq., according to my book, using substitution $$\frac{y}{x}=z(x)$$

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}=\frac{r^2}{(x(1+\frac{y}{x}))^2}=\frac{r^2}{x^2(1+\frac{y}{x})^2}$$

Our equation now becomes $$z+x\frac{\mathrm{d} z}{\mathrm{d} x}=\frac{r^2}{x^2(1+z)^2}$$ but this is not diff. eq. with separate variables?

"Differential equations of the form $$\frac{\mathrm{d} y}{\mathrm{d} x}=f(x,y)$$, where $$f(x,y)$$ is homogenous function (function is homogenous if $$f(tx,ty)=t^k f(x,y)$$) can be written in form $$\frac{\mathrm{d} y}{\mathrm{d} x}=F(\frac{y}{x})$$ and transformed to differential equation with separate variables using substitution $$\frac{y}{x}=z(x)$$."

Here is my differential equation and how I tried to solve it:

$$(x+y)^2 \frac{\mathrm{d} y}{\mathrm{d} x}=r^2$$

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}$$

$$f(x,y)=\frac{r^2}{(x+y)^2}$$

$$\rightarrow f(tx,ty)=\frac{r^2}{(tx+ty)^2}=\frac{r^2}{(t(x+y))^2}=\frac{r^2}{t^2(x+y)^2}=t^{-2}\frac{r^2}{(x+y)^2}$$ so our function is homogenous and we can solve our diff. eq., according to my book, using substitution $$\frac{y}{x}=z(x)$$

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{r^2}{(x+y)^2}=\frac{r^2}{(x(1+\frac{y}{x}))^2}=\frac{r^2}{x^2(1+\frac{y}{x})^2}$$

Our equation now becomes $$z+x\frac{\mathrm{d} z}{\mathrm{d} x}=\frac{r^2}{x^2(1+z)^2}$$ but this is not diff. eq. with separate variables?

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