JEE Main 2023 — Mathematics Calculus

Given,

$$(3{y}^{2}-5{x}^{2})ydx+2x({x}^{2}-{y}^{2})dy=0$$(3y2−5x2)ydx+2x(x2−y2)dy=0

$$\Rightarrow \frac{dy}{dx}=\frac{y(5{x}^{2}-3{y}^{2})}{2x({x}^{2}-{y}^{2})}$$⇒dxdy​=2x(x2−y2)y(5x2−3y2)​

This is homogeneous differential equation

Let $y=tx\Rightarrow \frac{dy}{dx}=t+x\frac{dt}{dx}$y=tx⇒dxdy​=t+xdxdt​

$$t+x\frac{dt}{dx}=\frac{(tx)(5{x}^{2}-3{t}^{2}{x}^{2})}{2x({x}^{2}-{t}^{2}{x}^{2})}=\frac{t(5-3{t}^{2})}{2(1-{t}^{2})}$$t+xdxdt​=2x(x2−t2x2)(tx)(5x2−3t2x2)​=2(1−t2)t(5−3t2)​

$$\Rightarrow x\frac{dt}{dx}=\frac{5t-3{t}^{3}}{2-2{t}^{2}}-t=\frac{3t-{t}^{3}}{2-2{t}^{2}}$$⇒xdxdt​=2−2t25t−3t3​−t=2−2t23t−t3​

$$\Rightarrow \frac{2(1-{t}^{2})}{3t-{t}^{3}}dt=\frac{1}{x}dx$$⇒3t−t32(1−t2)​dt=x1​dx

Taking integration on both sides we get,

$$\int \frac{2(1-{t}^{2})}{3t-{t}^{3}}dt=\int \frac{1}{x}dx......(1)$$∫3t−t32(1−t2)​dt=∫x1​dx......(1)

Now let $3t-{t}^{3}=z\Rightarrow 3(1-{t}^{2})dt=dz$3t−t3=z⇒3(1−t2)dt=dz

$$\Rightarrow (1-{t}^{2})dt=\frac{1}{3}dz$$⇒(1−t2)dt=31​dz

Substitute in equation $(1)$(1) we get,

$$\frac{2}{3}\int \frac{dz}{z}=\int \frac{1}{x}dx$$32​∫zdz​=∫x1​dx

$$\Rightarrow \frac{2}{3}\mathrm{ln}(z)=\mathrm{ln}x+\mathrm{ln}c$$⇒32​ln(z)=lnx+lnc

$$\Rightarrow {z}^{\frac{2}{3}}=cx\Rightarrow z={c}^{\frac{3}{2}}{x}^{\frac{3}{2}}$$⇒z32​=cx⇒z=c23​x23​

$$\Rightarrow 3t-{t}^{3}={c}^{\frac{3}{2}}{x}^{\frac{3}{2}}$$⇒3t−t3=c23​x23​

Now put $t=\frac{y}{x}$t=xy​

$$\Rightarrow \frac{3y}{x}-\frac{{y}^{3}}{{x}^{3}}={c}^{\frac{3}{2}}{x}^{\frac{3}{2}}$$⇒x3y​−x3y3​=c23​x23​

Given $y(1)=1$y(1)=1 when $x=1,y=1$x=1,y=1

$$\Rightarrow {c}^{\frac{3}{2}}=2$$⇒c23​=2

So, equation becomes $3\frac{y}{x}-\frac{{y}^{3}}{{x}^{3}}=2{x}^{\frac{3}{2}}$3xy​−x3y3​=2x23​

$$\Rightarrow 3{x}^{2}y-{y}^{3}=2{x}^{\frac{9}{2}}......(2)$$⇒3x2y−y3=2x29​......(2)

Put $x=2$x=2 in equation $(2)$(2),

$$12y(2)-{(y(2))}^{3}=2(2{)}^{\frac{9}{2}}=32\sqrt{2}$$12y(2)−(y(2))3=2(2)29​=322​

$$\therefore |{(y(2))}^{3}-12y(2)|=32\sqrt{2}$$∴∣(y(2))3−12y(2)∣=322​