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Find the solution of the differential equation $\displaystyle \left(\frac{\mathrm dy}{\mathrm dx}\right)^2-x\frac{\mathrm dy}{\mathrm dx}+y=0$.

$\bf{My\; Try:}$ Let $\displaystyle \frac{\mathrm dy}{\mathrm dx} = t\;,$ then the differential equation gets converted into $t^2-xt+y=0$.

So, its solution is given by $\displaystyle t=\frac{x\pm \sqrt{x^2-4y}}{2}$.

Hence, we get $$\frac{\mathrm dy}{\mathrm dx} = \frac{x\pm \sqrt{x^2-4y}}{2}$$

Now, how can I proceed after this? Please help me.

Integreek
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juantheron
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4 Answers4

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Here's another approach. Differentiating gives $y''\left(2y'-x\right)=0$, so $y''=0$ or $y'=\frac{x}{2}$. The former option gives $y=ax+b$ so $a^2-ax+ax+b=0$ and $b=-a^2$. The latter option gives $y=\tfrac{x^2}{4}+c$ so $\frac{x^2}{4}-\frac{x^2}{2}+\tfrac{x^2}{2}+c=0$, which work iff $c=0$. The solution is $y=ax-a^2$ or $y=\frac{x^2}{4}$.

J.G.
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4

If you rewrite your equation by $y=xy'-y'^2$, you can see that your equation is actually a Clairaut's diffrerential equation with a general solution $y=ax-a^2.$ This equation also has a singular solution that can be obtained by solving the set of equations $F(x,y,a)=0$ and $F_a(x,y,a)=0$ where $F(x,y,a)=ax-a^2-y$. Since $F_a=x-2a$, you get $y=\frac{x^2}{4}$ as observed by Michael.

Marko
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4

$$y'(x)^2-xy'(x)+y(x)=0\Longleftrightarrow$$ $$y(x)=-y'(x)^2+xy'(x)\Longleftrightarrow$$ $$y'(x)=xy''(x)+y'(x)-2y'(x)y''(x)\Longleftrightarrow$$ $$y'(x)=y'(x)+y''(x)\left(x-2y'(x)\right)\Longleftrightarrow$$ $$y''(x)\left(x-2y'(x)\right)=0$$

Now, solve them separately:

  • For the first one:

$$y''(x)=0\Longleftrightarrow$$ $$\int y''(x)\space\text{d}x=\int0\space\text{d}x\Longleftrightarrow$$ $$\int y'(x)\space\text{d}x=\text{C}_1\Longleftrightarrow$$

Substitute $y'(x)=\text{C}_1$ into $y(x)=xy'(x)-y'(x)^2$:

$$y(x)=\text{C}_1x-\text{C}_1^2$$

  • For the second one: $$x-2y'(x)=0\Longleftrightarrow$$ $$y'(x)=\frac{x}{2}\Longleftrightarrow$$

Substitute into $y(x)=xy'(x)-y'(x)^2$:

$$y(x)=\frac{x^2}{4}$$

So, finally we found that:

$$y(x)=\frac{x^2}{4}\space\space\space\space\space\space\space\text{or}\space\space\space\space\space\space\space y(x)=\text{C}_1x-\text{C}_1^2$$

Jan Eerland
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0

You are on the right track. From where you left off, the ODE can be reduced into a variable-separable form by rearranging it as follows and recognizing the differential of $\frac{x^2}4-y$:

$$\frac{\mathrm dy}{\mathrm dx}=\frac{x\pm\sqrt{x^2-4y}}2$$ $$\implies\mp\sqrt{\frac{x^2}4-y}\,\mathrm dx=\frac{x}2\mathrm dx-\mathrm dy$$ $$\implies\frac{\mathrm d\left(\frac{x^2}4-y\right)}{\sqrt{\frac{x^2}4-y}}=\mp\mathrm dx\tag{if $x^2-4y\ne0$}$$ $$\therefore\mp\sqrt{x^2-4y}=x+C, x^2=4y$$

The first condition can be simplified by squaring both sides:

$$\require{cancel}\cancel{x^2}-4y=\cancel{x^2}+2Cx+C^2$$ $$\implies y=\left(\frac{-C}2\right)x-\left(\frac{-C}2\right)^2$$ $$\therefore y=cx-c^2$$

Integreek
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