JEE Main 2022 — Mathematics Calculus

Given $(4+{x}^{2})dy-2x({x}^{2}+3y+4)dx=0$(4+x2)dy−2x(x2+3y+4)dx=0

$$({x}^{2}+4)\frac{dy}{dx}=2{x}^{3}+6xy+8x$$(x2+4)dxdy​=2x3+6xy+8x

$$({x}^{2}+4)\frac{dy}{dx}-6xy=2{x}^{3}+8x$$(x2+4)dxdy​−6xy=2x3+8x

$$\frac{dy}{dx}-\frac{6x}{{x}^{2}+4}y=\frac{2{x}^{3}+8x}{{x}^{2}+4}$$dxdy​−x2+46x​y=x2+42x3+8x​

This is of the form of linear differential equation

I.F. $\displaystyle ={e}^{-\int \frac{6x}{{x}^{2}+4}dx}={e}^{-3{\mathrm{log}}_{e}({x}^{2}+4)}$=e−∫x2+46x​dx=e−3loge​(x2+4)

$$={e}^{{\mathrm{log}}_{e}{({x}^{2}+4)}^{-3}}=\frac{1}{{({x}^{2}+4)}^{3}}$$=eloge​(x2+4)−3=(x2+4)31​

So the solution of the differential equation will be

$$y.\frac{1}{{({x}^{2}+4)}^{3}}=\int \frac{2{x}^{3}+8x}{{({x}^{2}+4)}^{3}({x}^{2}+4)}dx$$y.(x2+4)31​=∫(x2+4)3(x2+4)2x3+8x​dx

$$\frac{y}{{({x}^{2}+4)}^{3}}=\int \frac{2x({x}^{2}+4)}{{({x}^{2}+4)}^{3}({x}^{2}+4)}dx$$(x2+4)3y​=∫(x2+4)3(x2+4)2x(x2+4)​dx

Let ${x}^{2}+4=t$x2+4=t, $2xdx=dt$2xdx=dt

So $\displaystyle \frac{y}{{({x}^{2}+4)}^{3}}=\int \frac{dt}{{t}^{3}}$(x2+4)3y​=∫t3dt​

$$\frac{y}{{({x}^{2}+4)}^{3}}=\frac{-1}{2{({x}^{2}+4)}^{2}}+C$$(x2+4)3y​=2(x2+4)2−1​+C

Since the curve passes through origin $(0,0)$(0,0)

$$0=\frac{-1}{2\times 16}+C\Rightarrow C=\frac{1}{32}$$0=2×16−1​+C⇒C=321​

i.e. $\frac{y}{{({x}^{2}+4)}^{3}}=\frac{-1}{2{({x}^{2}+4)}^{2}}+\frac{1}{32}$(x2+4)3y​=2(x2+4)2−1​+321​

$$y=\frac{-({x}^{2}+4)}{2}+\frac{{({x}^{2}+4)}^{3}}{32}$$y=2−(x2+4)​+32(x2+4)3​

Hence $y(2)=-\frac{8}{2}+\frac{8\times 8\times 8}{32}=12$y(2)=−28​+328×8×8​=12