GATE2019 ME-2: 3

Question

The differential equation $\dfrac{dy}{dx}+4y=5$ is valid in the domain $0 \leq x \leq 1$ with $y(0)=2.25$. The solution of the differential equation is

• A. $y=e^{-4x}+5$
• B. $y=e^{-4x}+1.25$
• C. $y=e^{4x}+5$
• D. $y=e^{4x}+1.25$

Answer 1

If we have a differential equation $\frac{\mathrm{d} y}{\mathrm{d} x} + P(x)y = Q(x)$

where $P(x)$ and $Q(x)$ are functions in ‘$x$’.

Then solution of this differential equation is given by :-

$y(I.F.) = \int Q(x)(I.F.)\;dx + c$

where I.F. is integrating factor which is defined as $e^{\int P(x)\;dx}$ and ‘c‘ is arbitrary constant.

So, here, $P(x)=4$ and $Q(x)=5$

Now, I.F. = $e^{\int 4dx} = e^{4x}$

So, Solution of given differential equation is :-

$ye^{4x} = \int 5e^{4x}\;dx + c$

$\Rightarrow$ $ye^{4x} = \frac{5}{4}e^{4x}\; + c$

$\Rightarrow$ $y = \frac{5}{4}\; + ce^{-4x}$

Now, It is given that $y(0)=2.25$

So, $2.25=\frac{5}{4}+c*e^{-0}$

$\Rightarrow$ $c=1.00$

So, Solution of given differential equation will be :-

$y = \frac{5}{4}\; + 1*e^{-4x}$

$\Rightarrow$ $y = 1.25\; + e^{-4x}$

So, Answer is $(B)$

Reference :- https://en.wikipedia.org/wiki/Integrating_factor