JEE Main 2021 — Mathematics Calculus

Given that

$$x\phi (x)={\int }_{5}^{x}(3{t}^{2}-2{\phi }^{'}(t))dt$$xϕ(x)=∫5x​(3t2−2ϕ′(t))dt

$$\phi (0)=4,x>-2$$ϕ(0)=4,x>−2

Differentiating on both the sides and using Newton Leibnitz rule

$$\Rightarrow x{\phi }^{'}(x)+1\phi (x)=3{x}^{2}-2{\phi }^{'}(x)$$⇒xϕ′(x)+1ϕ(x)=3x2−2ϕ′(x)

$$\because \frac{d}{dx}(UV)=U\frac{dV}{dx}+V\frac{dU}{dx};\frac{d}{dx}{\int }_{f(x)}^{g(x)}h(x)dx={g}^{'}(x)h(g(x))-{f}^{'}(x)h(f(x))$$∵dxd​(UV)=UdxdV​+VdxdU​;dxd​∫f(x)g(x)​h(x)dx=g′(x)h(g(x))−f′(x)h(f(x))

$$\Rightarrow (x+2){\phi }^{'}(x)+\phi (x)=3{x}^{2}$$⇒(x+2)ϕ′(x)+ϕ(x)=3x2

It is a linear differential equation in the form of $\frac{dy}{dx}+Py=Q$dxdy​+Py=Q

$$\Rightarrow I.F.={e}^{\int (\frac{1}{x+2})dx}=x+2$$⇒I.F.=e∫(x+21​)dx=x+2

$$\Rightarrow \phi (x)\cdot (x+2)=\int (x+2)(\frac{3{x}^{2}}{x+2})dx$$⇒ϕ(x)⋅(x+2)=∫(x+2)(x+23x2​)dx

$$\Rightarrow \phi (x)\cdot (x+2)={x}^{3}+c$$⇒ϕ(x)⋅(x+2)=x3+c

Given that $\phi (0)=4$ϕ(0)=4

$$\Rightarrow c=8$$⇒c=8

$$\Rightarrow \phi (2)=\frac{8+8}{4}=4$$⇒ϕ(2)=48+8​=4