Let y=y(x),y0, be a solution curve of the differential equation (1+x^2)dy=y(x-y)dx.

Question

Let $y=y(x),y>0$, be a solution curve of the differential equation $(1+{x}^{2})dy=y(x-y)dx$. If $y(0)=1$ and $y(2\sqrt{2})=\beta $, then

Accepted Answer

$\begin{gathered} (1+{x}^{2})dy=y(x-y)dx \ y(0)=1\cdot y(2\sqrt{2})=\beta \ \frac{dy}{dx}=\frac{yx-{y}^{2}}{1+{x}^{2}} \ \frac{dy}{dx}+y(\frac{-x}{1+{x}^{2}})=(\frac{-1}{1+{x}^{2}}){y}^{2} \ \frac{1}{{y}^{2}}\frac{dy}{dx}+\frac{1}{y}(\frac{-x}{1+{x}^{2}})=\frac{-1}{1+{x}^{2}} \ \frac{1}{y}=t\frac{-1}{{y}^{2}}\frac{dy}{dx}=\frac{dt}{dx} \end{gathered}$

$\frac{dt}{dx}+t\frac{x}{1+{x}^{2}}=\frac{1}{1+{x}^{2}}$

$={e}^{\int \frac{x}{1+{x}^{2}}dx}={e}^{\frac{1}{2}\ln (1+{x}^{2})}=\sqrt{1+{x}^{2}}$

$t\sqrt{1+{x}^{2}}=\int \frac{\sqrt{1+{x}^{2}}}{1+{x}^{2}}dx$

$\begin{gathered} \frac{1}{y}\sqrt{1+{x}^{2}}=\int \frac{1}{\sqrt{1+{x}^{2}}}dx \ \frac{1}{y}\sqrt{1+{x}^{2}}=\ln (x+\sqrt{{x}^{2}+1})+C \ \because y(0)=1\Rightarrow C=1 \ \frac{1}{y}\sqrt{1+{x}^{2}}=\ln (x+\sqrt{{x}^{2}+1})+1 \ y=2\sqrt{2} \ \frac{3}{y}=\ln |2\sqrt{2}+3|+1 \end{gathered}$

$\begin{gathered} y=\beta =\frac{3}{1+\ln |2\sqrt{2}+3|} \ \Rightarrow 3{\beta }^{-1}=1+\ln |2\sqrt{2}+3| \end{gathered}$

Isolating the term ${e}^{3{\beta }^{-1}}$, we get :

${e}^{3{\beta }^{-1}}={e}^{1+\ln |2\sqrt{2}+3|}.$

This can be simplified using the rule ${e}^{a+b}={e}^{a}{e}^{b}$ to :

${e}^{3{\beta }^{-1}}=e\cdot {e}^{\ln |2\sqrt{2}+3|}.$

Since ${e}^{\ln (x)}=x$ for any $x$, this simplifies to :

${e}^{3{\beta }^{-1}}=e|2\sqrt{2}+3|.$

Using the given value of $\beta $, which is $\frac{3}{1+\ln |2\sqrt{2}+3|}$, we find :

${e}^{3{\beta }^{-1}}=e|2\sqrt{2}+3|=e(2\sqrt{2}+3),$

Since $2\sqrt{2}+3$ is positive and so the absolute value does not affect the result.

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