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# GATE-2012 ECE Q12 (math)

by on December 28, 2012

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

## 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,

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

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

## 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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