JEE Main 2023 — Mathematics Calculus

Given,

$$(1+{x}^{2})dy=y(x-y)dx$$(1+x2)dy=y(x−y)dx

$$\Rightarrow \frac{dy}{dx}=\frac{x}{1+{x}^{2}}y-\frac{{y}^{2}}{1+{x}^{2}}$$⇒dxdy​=1+x2x​y−1+x2y2​

$$\Rightarrow \frac{-1}{{y}^{2}}\frac{dy}{dx}+\frac{x}{1+{x}^{2}}\cdot \frac{1}{y}=\frac{1}{1+{x}^{2}}$$⇒y2−1​dxdy​+1+x2x​⋅y1​=1+x21​

Now let $\frac{1}{y}=t$y1​=t, we get,

$$\frac{-1}{{y}^{2}}{y}^{'}=\frac{dt}{dx}$$y2−1​y′=dxdt​

So, the equation becomes,

$$\frac{dt}{dx}+\frac{x}{1+{x}^{2}}t=\frac{1}{1+{x}^{2}}$$dxdt​+1+x2x​t=1+x21​

Now finding, $\displaystyle \mathrm{IF}={e}^{\int \frac{x}{1+{x}^{2}}dx}=\sqrt{1+{x}^{2}}$IF=e∫1+x2x​dx=1+x2​

So, solution of the differential equation is given by,

$$t\sqrt{1+{x}^{2}}=\int \frac{\sqrt{1+{x}^{2}}}{1+{x}^{2}}dx$$t1+x2​=∫1+x21+x2​​dx

$$\Rightarrow \frac{1}{y}\sqrt{1+{x}^{2}}=\int \frac{1}{\sqrt{1+{x}^{2}}}dx$$⇒y1​1+x2​=∫1+x2​1​dx

$$\Rightarrow \frac{1}{y}\sqrt{1+{x}^{2}}=\mathrm{ln}(x+\sqrt{{x}^{2}+1})+C$$⇒y1​1+x2​=ln(x+x2+1​)+C

$$\because y(0)=1\Rightarrow C=1$$∵y(0)=1⇒C=1

$$\Rightarrow \frac{1}{y}\sqrt{1+{x}^{2}}=\mathrm{ln}(x+\sqrt{{x}^{2}+1})+1$$⇒y1​1+x2​=ln(x+x2+1​)+1

Now for $x=2\sqrt{2},y=\beta$x=22​,y=β we get,

$$\Rightarrow \frac{3}{\beta }=\mathrm{ln}|2\sqrt{2}+3|+1$$⇒β3​=ln∣22​+3∣+1

$$\Rightarrow \beta =\frac{3}{1+\mathrm{ln}|2\sqrt{2}+3|}$$⇒β=1+ln∣22​+3∣3​

$$\Rightarrow 3{\beta }^{-1}=1+\mathrm{ln}|2\sqrt{2}+3|$$⇒3β−1=1+ln∣22​+3∣

$$\Rightarrow {e}^{3\beta -1}=e\cdot {e}^{\mathrm{ln}(3+2\sqrt{2})}$$⇒e3β−1=e⋅eln(3+22​)

$$\Rightarrow {e}^{3\beta -1}=e(3+2\sqrt{2})$$⇒e3β−1=e(3+22​)