JEE Main 2024 — Mathematics Calculus

Given: $(t+1)dx=(2x+{(t+1)}^{4})dt$(t+1)dx=(2x+(t+1)4)dt

$$\Rightarrow \frac{dx}{dt}=\frac{2x+{(t+1)}^{4}}{(t+1)}$$⇒dtdx​=(t+1)2x+(t+1)4​

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

$$\Rightarrow \mathrm{IF}={e}^{\int \frac{-2}{t+1}dt}$$⇒IF=e∫t+1−2​dt

$$\Rightarrow \mathrm{IF}={e}^{-2\mathrm{log}|t+1|}$$⇒IF=e−2log∣t+1∣

$$\Rightarrow \mathrm{IF}=\frac{1}{{(t+1)}^{2}}$$⇒IF=(t+1)21​

Solution of the differential equations is given by,

$$\Rightarrow x\times \frac{1}{{(t+1)}^{2}}=\int (t+1)dt$$⇒x×(t+1)21​=∫(t+1)dt

$$\Rightarrow \frac{x}{{(t+1)}^{2}}=\frac{{t}^{2}}{2}+t+c$$⇒(t+1)2x​=2t2​+t+c

It is given that $x(0)=2$x(0)=2

$$\Rightarrow \frac{2}{{(0+1)}^{2}}=\frac{0}{2}+0+c$$⇒(0+1)22​=20​+0+c

$$\Rightarrow c=2$$⇒c=2

$$\Rightarrow \frac{x}{{(t+1)}^{2}}=\frac{{t}^{2}}{2}+t+2$$⇒(t+1)2x​=2t2​+t+2

Putting $t=1$t=1,

$$\Rightarrow \frac{x}{{(1+1)}^{2}}=\frac{{1}^{2}}{2}+1+2$$⇒(1+1)2x​=212​+1+2

$$\Rightarrow \frac{x}{4}=\frac{1}{2}+3$$⇒4x​=21​+3

$$\Rightarrow x=2+12$$⇒x=2+12

$$\Rightarrow x=14$$⇒x=14