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GATE-2012 ECE Q12 (math)

Krishna Sankar12 years ago2 years ago01 mins

Question 12 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q12. 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) ${x=\frac{t}{2}}$

Solution

From the product rule used to find the derivative of product of two or more functions,

${\frac{d}{dx}(f.g)=f\frac{dg}{dx}+g\frac{df}{dx}}$

Applying this to the above equation, we can be seen that,

$\frac{d}{dt}(t.x)=t\frac{dx}{dt}+x\frac{dt}{dt}=t\frac{dx}{dt}+x$

Plugging this in and integrating both sides,

$\begin{array}t\frac{dx}{dt}+x&=&t\\\frac{d}{dt}(t.x)&=&t\\\int\frac{d}{dt}(t.x)&=&\int t\\tx&=&\frac{t^2}{2}+C\end{array}$ .

Using the initial condition ${x(1)=0.5}$ , we can solve for the unknown $C$ , i.e.
$\begin{array}t.x&=&\frac{t^2}{2}+C\\1*0.5&=&\frac{1^2}{2}+C\\C&=&0\end{array}$ .

So the solution to the differential equation is,

${x=\frac{t}{2}}$

Based on the above, the right choice is (D) ${x=\frac{t}{2}}$ .

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Wiki entry on Product rule

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