If x=x(t) is the solution of the differential | vedclass

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(A) Given the differential equation: $(t+1) dx = (2x + (t+1)^4) dt$. Rearranging into the standard linear form $\frac{dx}{dt} + P(t)x = Q(t)$: $\frac{

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$14$

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(A) Given the differential equation: $(t+1) dx = (2x + (t+1)^4) dt$. Rearranging into the standard linear form $\frac{dx}{dt} + P(t)x = Q(t)$: $\frac{dx}{dt} = \frac{2x + (t+1)^4}{t+1} = \frac{2x}{t+1} + (t+1)^3$. $\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$. The integrating factor $I.F. = e^{\int -\frac{2}{t+1} dt} = e^{-2 \ln(t+1)} = (t+1)^{-2} = \frac{1}{(t+1)^2}$. Multiplying both sides by $I.F.$: $\frac{d}{dt} \left( \frac{x}{(t+1)^2} \right) = (t+1)^3 \cdot \frac{1}{(t+1)^2} = (t+1)$. Integrating both sides with respect to $t$: $\frac{x}{(t+1)^2} = \int (t+1) dt = \frac{(t+1)^2}{2} + C$. Using the initial condition $x(0) = 2$: $\frac{2}{(0+1)^2} = \frac{(0+1)^2}{2} + C \Rightarrow 2 = \frac{1}{2} + C \Rightarrow C = \frac{3}{2}$. Thus,the solution is $x = \frac{(t+1)^4}{2} + \frac{3}{2}(t+1)^2$. For $t=1$: $x(1) = \frac{(1+1)^4}{2} + \frac{3}{2}(1+1)^2 = \frac{16}{2} + \frac{3}{2}(4) = 8 + 6 = 14$.

If $x=x(t)$ is the solution of the differential equation $(t+1) dx = (2x + (t+1)^4) dt$ with the initial condition $x(0) = 2$,then $x(1)$ equals:

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