Grimanijd

2022-07-13

I have difficulties to solve these two differential equations:

1) ${y}^{\prime}(x)=\frac{x-y(x)}{x+y(x)}$ with the initial condition $y(1)=1$ .I'm arrived to prove that

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

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) ${y}^{\prime}(x)=\frac{2y(x)-x}{2x-y(x)}$. 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 $c$ ?

1) ${y}^{\prime}(x)=\frac{x-y(x)}{x+y(x)}$ with the initial condition $y(1)=1$ .I'm arrived to prove that

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

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) ${y}^{\prime}(x)=\frac{2y(x)-x}{2x-y(x)}$. 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 $c$ ?

zlepljalz2

Beginner2022-07-14Added 22 answers

${y}^{\prime}(y+x)=x-y$

${y}^{\prime}x+y=x-{y}^{\prime}y$

$(xy{)}^{\prime}=x-\frac{1}{2}({y}^{2}{)}^{\prime}$

Integrate

$xy=\frac{1}{2}{x}^{2}-\frac{1}{2}{y}^{2}+C$

${y}^{2}-{x}^{2}+2xy=C$

Evaluate the constant:

$y(1)=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=2$

$(y+x{)}^{2}=2({x}^{2}+1)$

Finally,

$\overline{){\displaystyle y(x)=\sqrt{2({x}^{2}+1)}-x}}$

You are on the right track

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

$y=x(\sqrt{2-\frac{{e}^{-2c}}{{x}^{2}}}-1)$

Note that ${e}^{-2c}=k$

$y=\sqrt{2{x}^{2}-k}-x$

$y=\sqrt{2{x}^{2}+2}-x$

${y}^{\prime}x+y=x-{y}^{\prime}y$

$(xy{)}^{\prime}=x-\frac{1}{2}({y}^{2}{)}^{\prime}$

Integrate

$xy=\frac{1}{2}{x}^{2}-\frac{1}{2}{y}^{2}+C$

${y}^{2}-{x}^{2}+2xy=C$

Evaluate the constant:

$y(1)=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=2$

$(y+x{)}^{2}=2({x}^{2}+1)$

Finally,

$\overline{){\displaystyle y(x)=\sqrt{2({x}^{2}+1)}-x}}$

You are on the right track

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

$y=x(\sqrt{2-\frac{{e}^{-2c}}{{x}^{2}}}-1)$

Note that ${e}^{-2c}=k$

$y=\sqrt{2{x}^{2}-k}-x$

$y=\sqrt{2{x}^{2}+2}-x$

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