JEE Main 2023 — Mathematics Calculus

Given the solution of differential equation,

$$\alpha x={e}^{{x}^{\beta }\cdot {y}^{\gamma }}......(1)$$αx=exβ⋅yγ......(1)

Now calculating the solution of differential equation,

$$2{x}^{2}y\frac{dy}{dx}=1-x{y}^{2}$$2x2ydxdy​=1−xy2

Let ${y}^{2}=t\Rightarrow 2\frac{dy}{dx}=\frac{dt}{dx}$y2=t⇒2dxdy​=dxdt​

$$\therefore {x}^{2}\frac{dt}{dx}=1-xt$$∴x2dxdt​=1−xt

$$\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{t}{x}=\frac{1}{{x}^{2}}.....(2)$$dxdt​+xt​=x21​.....(2)

Above differential equation is linear in $t$t

We know solutions of differential equation of the form $\frac{dy}{\mathrm{dx}}+Py=Q$dxdy​+Py=Q is given by,

$$y.{e}^{\int Pdx}=\int Q.{e}^{\int Pdx}+C({e}^{\int Pdx}=I.F.)$$y.e∫Pdx=∫Q.e∫Pdx+C(e∫Pdx=I.F.)

Now calculating, $\displaystyle I.F.={e}^{\int \frac{1}{x}dx}={e}^{\mathrm{ln}x}=x$I.F.=e∫x1​dx=elnx=x

Therefore, the solution of equation $(2)$(2) is given by,

$$t.x=\int \frac{1}{{x}^{2}}\cdot xdx+C$$t.x=∫x21​⋅xdx+C

$$\Rightarrow {y}^{2}\cdot x=\mathrm{ln}x+C$$⇒y2⋅x=lnx+C

$$\because y(2)=\sqrt{\mathrm{ln}2}$$∵y(2)=ln2​

$$\therefore 2.\mathrm{ln}2=\mathrm{ln}2+C\Rightarrow C=\mathrm{ln}2$$∴2.ln2=ln2+C⇒C=ln2

Hence, solution is $x{y}^{2}=\mathrm{ln}x+\mathrm{ln}2$xy2=lnx+ln2

$$\Rightarrow x{y}^{2}=\mathrm{ln}2x$$⇒xy2=ln2x

$$\Rightarrow 2x={e}^{x\cdot {y}^{2}}......(3)$$⇒2x=ex⋅y2......(3)

On comparing equations $(1)$(1) and $(3)$(3) we get,

$$\alpha =2,\beta =1,\gamma =2$$α=2,β=1,γ=2

$$\therefore \alpha +\beta -\gamma =1$$∴α+β−γ=1