Question

With initial condition x(1) = 0.5, the solution of the differential equation, t$\frac{dx}{dt}+x=t$ is

• A. $x=t-\frac{1}{2}$
• B. $x=t^2-\frac{1}{2}$
• C. $x=\frac{t^2}{2}$
• D. $x=\frac{t}{2}$

Answer 1

The correct option is D $x=\frac{t}{2}$

$t\frac{dx}{dt}+x=t,x\left(1\right)=\frac{1}{2}$

$\frac{dx}{dt}+\frac{1}{t}x=1$

which is a linear differential equation

$I.F=e^{\int \frac{1}{t}dt}=e^{logt}=t$

Solution is
$x.(I.F)=\int 1.\left(IF\right)dt+C$

$x\times t=\int t.1dt+c$

$tx=\frac{t^2}{2}+c$

Given that x(1) = $\frac{1}{2}$

$\Rightarrow 1\left(\frac{1}{2}\right)=\frac{1}{2}+c\Rightarrow c=0$

So, tx = $\frac{t^2}{2}\Rightarrow x=\frac{t}{2}$