Question

Let $y'(x) + y(x)g'(x) = g(x)g'(x),y(0) = 0,x \in \mathbb{R}$, where $f'(x)$ denotes $\dfrac{{df(x)}}{{dx}}$ and $g(x)$ is a given non – constant differentiable function on $\mathbb{R}$ with $g(0) = g(2) = 0$. Then the value of $y(2)$ is:
A. 0
B. 1
C. 2
D. 3

Answer 1

We will use the integration factor for the linear differential equation and thus get the value of y(x) in terms of g(x) and then put in x = 0 to find the constant and then put in x = 2 to get the required value.

Complete step-by-step solution:
We are given the linear differential equation as:-
$\Rightarrow y'(x) + y(x)g'(x) = g(x)g'(x)$
We can write the above equation as follows:-
$\Rightarrow \dfrac{{dy}}{{dx}} + y(x)g'(x) = g(x)g'(x)$
Now, if we compare this to the general linear differential equation which is $\dfrac{{dy}}{{dx}} + Py = Q$ where P and Q are functions of x only.
We can clearly see that P(x) = g’(x) and Q(x) = g’(x).g(x)
Now, we know that the equation $\dfrac{{dy}}{{dx}} + Py = Q$ has the integrating factor of ${e^{\int {Pdx} }}$.
And, then the solution of this equation is given by the following expression:-
$\Rightarrow y(x) \times I.F. = \int {Q \times I.F.}$
Therefore, the equation $\dfrac{{dy}}{{dx}} + y(x)g'(x) = g(x)g'(x)$ will have the integrating factor of ${e^{\int {g'(x)dx} }}$.
$\Rightarrow$ Integrating Factor = ${e^{g(x)}}$
Therefore, its solution will be:-
$\Rightarrow y(x) \times {e^{g(x)}} = \int {g(x)g'(x){e^{g(x)}}dx}$
Let us assume that g(x) = t which implies that g’(x)dx = dt
$\Rightarrow y(x){e^t} = \int {t{e^t}dt}$
Now, we will use the formula: $\int {\left( {u.v} \right)dx} = u\int {vdx - \int {\left( {\dfrac{d}{{dx}}\left( u \right)\int {vdx} } \right)} } dx$
So, we will then obtain:-
$\Rightarrow y(x){e^t} = t{e^t} - \int {{e^t}dt}$
Now, we also know that integration of exponential function is now altered.
So, we will now obtain:-
$\Rightarrow y(x){e^t} = t{e^t} - {e^t} + C$
We can rewrite it as:-
$\Rightarrow y(x) = t - 1 + C{e^{ - t}}$
Substituting g(x) = t back, we will obtain:-
$\Rightarrow y(x) = g(x) - 1 + C{e^{ - g(x)}}$ ……………….(1)
Now, let us put x = 0 in equation (1), we will then get:-
$\Rightarrow y(0) = g(0) - 1 + C{e^{ - g(0)}}$
Since, we are given that y(0) = 0 and g(0) = 0
$\Rightarrow 0 = 0 - 1 + C$
$\Rightarrow C = 1$
Putting this in equation (1), we will get:-
$\Rightarrow y(x) = g(x) - 1 + {e^{ - g(x)}}$ ……………..(2)
Now, let us put x = 2 in equation (2), we will then get:-
$\Rightarrow y(2) = g(2) - 1 + {e^{ - g(2)}}$
Since, we are given that g(2) = 0
$\Rightarrow y(2) = 0 - 1 + {e^{ - 0}}$
$\Rightarrow y(2) = - 1 + 1 = 0$

Hence, the correct answer is option (A).

Note: The students must note that it might be intriguing for you to use Rolle’s theorem on it because of the given condition on the function g(x) which resemble a lot like it but eventually when you come to y(x), you will have to apply converse of Rolle’s theorem, which is not necessarily always true for the same points. Therefore, it is not applicable here.