If x=x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)^4)dt,x(0)=2, then…

Question

If $x=x(t)$ is the solution of the differential equation $(t+1)dx=(2x+(t+1{)}^{4})dt,x(0)=2$, then, $x(1)$ equals \_\_\_\_\_\_\_\_\_.

Answer: 14

Accepted Answer

$\begin{gathered} (t+1)dx=(2x+(t+1{)}^{4})dt \ \frac{dx}{dt}=\frac{2x+(t+1{)}^{4}}{t+1} \ \frac{dx}{dt}-\frac{2x}{t+1}=(t+1{)}^{3} \end{gathered}$

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

$\begin{gathered} \frac{x}{(t+1{)}^{2}}=\int \frac{1}{(t+1{)}^{2}}(t+1{)}^{3}dt+C \end{gathered}$

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

Now $x(0)=2$

$\begin{gathered} \Rightarrow C=2 \ 	herefore x=(\frac{{t}^{2}}{2}+t+2)(t+1{)}^{2} \ x(1)=(\frac{1}{2}+1+2)(1+1{)}^{2} \ =\frac{7}{2}	imes 4=14 \end{gathered}$

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