JEE Main 2022 — Mathematics Calculus

Given

$$\frac{dy}{dx}+{e}^{x}({x}^{2}-2)y=({x}^{2}-2x)({x}^{2}-2){e}^{2x}$$dxdy​+ex(x2−2)y=(x2−2x)(x2−2)e2x

It is linear differential equation so,

$$IF={e}^{\int ({x}^{2}-2){e}^{x}dx}$$IF=e∫(x2−2)exdx

$\displaystyle ={e}^{{x}^{2}{e}^{x}-2\int x{e}^{x}dx-2{e}^{x}}$=ex2ex−2∫xexdx−2ex $={e}^{{x}^{2}{e}^{x}-2[x{e}^{x}-{e}^{x}]-2{e}^{x}}$=ex2ex−2[xex−ex]−2ex

$$IF={e}^{({x}^{2}-2x){e}^{x}}$$IF=e(x2−2x)ex

Now solution is given by

$$y\times {e}^{({x}^{2}-2x){e}^{x}}=\int {e}^{({x}^{2}-2x){e}^{x}}\times ({x}^{2}-2x)({x}^{2}-2){e}^{2x}$$y×e(x2−2x)ex=∫e(x2−2x)ex×(x2−2x)(x2−2)e2x

$$=\int {e}^{({x}^{2}-2x){e}^{x}}\times ({x}^{2}-2x){e}^{x}({x}^{2}-2){e}^{x}$$=∫e(x2−2x)ex×(x2−2x)ex(x2−2)ex

Now let $({x}^{2}-2x){e}^{x}=t$(x2−2x)ex=t

We get ${e}^{x}({x}^{2}-2x)+{e}^{x}(2x-2)dx=dt$ex(x2−2x)+ex(2x−2)dx=dt

$${e}^{x}({x}^{2}-2x+2x-2)dx=dt$$ex(x2−2x+2x−2)dx=dt

$${e}^{x}({x}^{2}-2)dx=dt$$ex(x2−2)dx=dt

$$y\times {e}^{({x}^{2}-2x){e}^{x}}=\int {e}^{t}\times t\times dt$$y×e(x2−2x)ex=∫et×t×dt

$$y\times {e}^{({x}^{2}-2x){e}^{x}}={e}^{t}(t-1)+c$$y×e(x2−2x)ex=et(t−1)+c

$$y\times {e}^{({x}^{2}-2x){e}^{x}}={e}^{({x}^{2}-2x){e}^{x}}(({x}^{2}-2x){e}^{x}-1)+c$$y×e(x2−2x)ex=e(x2−2x)ex((x2−2x)ex−1)+c

Now at $x=0y=0$x=0y=0

$$0\times {e}^{0}={e}^{0}(0x{e}^{x}-1)+c$$0×e0=e0(0xex−1)+c

$0=-1+c$0=−1+c $c=1$c=1

Putting the value of $c=1$c=1 we get the equation as

$$\Rightarrow y\times {e}^{({x}^{2}-2x){e}^{x}}={e}^{({x}^{2}-2x){e}^{x}}(({x}^{2}-2x){e}^{x}-1)+1$$⇒y×e(x2−2x)ex=e(x2−2x)ex((x2−2x)ex−1)+1

Now putting $x=2$x=2 in equation we get

$$y\times {e}^{(4-4){e}^{2}}={e}^{(4-4){e}^{2}}((4-4){e}^{2}-1)+1$$y×e(4−4)e2=e(4−4)e2((4−4)e2−1)+1

$$\Rightarrow y\times {e}^{0}={e}^{0}((0){e}^{2}-1)+1$$⇒y×e0=e0((0)e2−1)+1

$$\Rightarrow y=-1+1=0$$⇒y=−1+1=0