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QJEEM24Ma17f5: JEE Main 2024 Mathematics - Class 12 JEEM24Ma17f5 Solution: If is the solution of the differential equation , then, equals…

If x=x(t)x=x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)4)dt,x(0)=2(t+1) \mathrm{d} x=\left(2 x+(t+1)^4\right) \mathrm{dt}, x(0)=2 , then, x(1)x(1) equals _________.

Step-by-Step Verified Solution

amp;(t+1)dx=(2x+(t+1)4)dtamp;dxdt=2x+(t+1)4t+1amp;dxdt−2xt+1=(t+1)3\begin{aligned} & (\mathrm{t}+1) \mathrm{dx}=\left(2 \mathrm{x}+(\mathrm{t}+1)^4\right) \mathrm{dt} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2 \mathrm{x}+(\mathrm{t}+1)^4}{\mathrm{t}+1} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}-\frac{2 \mathrm{x}}{\mathrm{t}+1}=(\mathrm{t}+1)^3\end{aligned}

I.F=e−∫2t+1dt=e−2ln⁡(t+1)=1(t+1)2I.F=e^{-\int \frac{2}{t+1} d t}=e^{-2 \ln (t+1)}=\frac{1}{(t+1)^2}

amp;x(t+1)2=∫1(t+1)2(t+1)3dt+C\begin{aligned} & \frac{x}{(t+1)^2}=\int \frac{1}{(t+1)^2}(t+1)^3 d t+C\end{aligned}

x(t+1)2=t22+t+C\frac{x}{(t+1)^2}=\frac{t^2}{2}+t+C

Now x(0)=2x(0)=2

amp;⇒C=2amp;∴x=(t22+t+2)(t+1)2amp;x(1)=(12+1+2)(1+1)2amp;=72×4=14\begin{aligned}& \Rightarrow C=2 \\\\& \therefore x=\left(\frac{t^2}{2}+t+2\right)(t+1)^2 \\\\& x(1)=\left(\frac{1}{2}+1+2\right)(1+1)^2 \\\\& =\frac{7}{2} \times 4=14\end{aligned}

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