Resolution of first order differential equation

Question

I have difficulties to solve these two differential equations:

1) $ y'(x)=\frac{x-y(x)}{x+y(x)} $ with the initial condition $ y(1)=1 $ .I'm arrived to prove that $$ y=x(\sqrt{2-e^{-2(\ln x+c)}}-1) $$ but I don't know if it's correct. If it's right, how do I find the constant $ c $? Because WolframAlpha says that the solution is $ y(x)=\sqrt{2}\sqrt{x^2+1}-x $.

2) $ y'(x)=\frac{2y(x)-x}{2x-y(x)} $ . I'm arrived to prove that $ \frac{z-1}{(z+1)^3}=e^{2c}x^{2} $ but I don't know if it's correct. If it's right, how do I explain $ z $ to substitute it in $ y=xz $? Then, how do I find the constant $ c $ ?

Thanks for any help!

Answer 1

$$y'(y+x)=x-y$$ $$y'x+y=x-y'y$$ $$(xy)'=x-\frac 12 (y^2)'$$ Integrate $$xy=\frac 12 x^2-\frac 12y^2+C$$ $$y^2-x^2+2xy=C$$ Evaluate the constant : $$y(1)=1 \implies C=2$$ $$(y+x)^2=2(x^2+1)$$ Finally, $$\boxed {y(x)=\sqrt {2(x^2+1)}-x}$$

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You are on the right track $$y=x(\sqrt{2-e^{-2(\ln x+c)}}-1)$$ $$y=x(\sqrt{2-\frac {e^{-2c}}{x^2}}-1)$$ Note that $e^{-2c}=k$ $$y=\sqrt{2x^2-k}-x$$ $$y=\sqrt{2x^2+2}-x$$

Answer 2

HINT

First of all, notice that \begin{align*} y^{\prime} = \frac{x-y}{x+y} \Longleftrightarrow y^{\prime} = \frac{1 - \frac{y}{x}}{1 + \frac{y}{x}} \end{align*}

Then, if we make $y = ux$, we get: \begin{align*} u + u^{\prime}x = \frac{1-u}{1+u} \Longleftrightarrow u^{\prime}x = \frac{1-u}{1+u} - u = \frac{1-2u-u^{2}}{1+u} \Longleftrightarrow \left[\frac{1+u}{2 - (1+u)^{2}}\right]u^{\prime} = \frac{1}{x} \end{align*}

The same trick applies to the second case. Can you proceed from here?

EDIT

If we make $y = ux$, we get: \begin{align*} y^{\prime} = \frac{2y-x}{2x-y} & \Longleftrightarrow u + u^{\prime}x = \frac{2u-1}{2 - u}\\ & \Longleftrightarrow u^{\prime}x = \frac{2u-1}{2 - u} - u = \frac{u^{2}-1}{u-2}\\ & \Longleftrightarrow \left[\frac{u-2}{u^{2}-1}\right]u^{\prime} = \frac{1}{x} \end{align*}

Answer 3

You may write your equation like $dy(x+y)=(x-y)dx $ $\Rightarrow$ $ydy+xdy=xdx-ydy$ $\Rightarrow$

$ydy+xdy+ydy=xdx$ $\Rightarrow$ $(y^2/2)'+(xy)'=(x^2/2)'$ $\Rightarrow$ $y^2+2xy=x^2$ $\Rightarrow$ $(y+x)^2=2x^2+c$

$\Rightarrow$ $y=\pm\sqrt{2x^2+c}-x$ . Now for the initial condition $y(1)=1$, only the positive root works. You may do the same for the second. Hint: You can always check your answer by differentiating your solution and substitute it to the initial equation.

Answer 4

Sorry I have not seen your second question

Your result is correct. Don't forget to substitute $e^c=K$

I susbstitued $y=tx$ and I got this $$\frac {t-1}{(t+1)^2}=Kx^2$$ Then substitute back $t=y/x$ $$\frac {y-x}{(y+x)^3}=K$$ You didn't give the initial condition so lets consider $y(x_0)=y_0$ $$K=\frac {y_0-x_0}{(y_0+x_0)^3}$$ Therefore $$\frac {y-x}{(y+x)^3}=\frac {y_0-x_0}{(y_0+x_0)^3}$$