Question

# Let y (x) be the solution of the differential equation $x\log x\dfrac{dy}{dx}+y=2x\log x$, $x\ge 1$. Then y (e) is equal to,(a). e(b). 0(c).2(d). 2e

Hint: Convert the given equation in the $\dfrac{dy}{dx}+P(x)\times y=Q(x)$ form and then use the formula $y\times \left( Integration\text{ }Factor \right)=\int{Q\left( x \right)}\times \left( Integration\text{ }Factor \right)dx$ + C you will get the y(x). Then use the boundary condition of $x\ge 1$ i.e. x = 1 you will get ‘C’ and then put x = e to get the value of y(e).

Complete step-by-step solution -
To solve the given problem we will write the given equation first,
$x\log x\dfrac{dy}{dx}+y=2x\log x$ ……………………………………………… (i)
Now to simplify the equation we will divide it by ‘xlogx’ on both sides of the equation, therefore we will get,
$\therefore \dfrac{x\log x\dfrac{dy}{dx}+y}{x\log x}=\dfrac{2x\log x}{x\log x}$
By separating the denominator we will get,
$\therefore \dfrac{x\log x\dfrac{dy}{dx}}{x\log x}+\dfrac{y}{x\log x}=\dfrac{2x\log x}{x\log x}$
$\therefore \dfrac{dy}{dx}+\dfrac{1}{x\log x}\times y=2$
If we compare the above equation with $\dfrac{dy}{dx}+P(x)\times y=Q(x)$ we will get,
$P(x)=\dfrac{1}{x\log x}$ and Q(x) = 2 ……………………………………………… (1)
Now to proceed further in the solution we should know the concept given below,
Concept:
If, $\dfrac{dy}{dx}+P(x)\times y=Q(x)$ then the solution for the differential equation is given by, $y\times \left( Integration\text{ }Factor \right)=\int{Q\left( x \right)}\times \left( Integration\text{ }Factor \right)dx$ + C ……………………… (2)
Where, Integration Factor = ${{e}^{\int{P\left( x \right)}\times dx}}$
To find the solution of the given differential equation we have to find the integrating factor first, therefore we will get,
Integration Factor = ${{e}^{\int{P\left( x \right)}\times dx}}$
If we put the value of equation (1) in the above equation we will get,
Therefore, Integration Factor $={{e}^{\int{\dfrac{1}{x\log x}}\times dx}}$
Now put, log x = t ……………………………………………………………. (3)
Therefore, $\dfrac{1}{x}dx=dt$
If we substitute the above values in equation we will get,
Therefore, Integration Factor $={{e}^{\int{\dfrac{dt}{t}}}}$
Now to proceed further in the solution we should know the formula given below,
Formula:
$\int{\dfrac{1}{x}}\times dx=\ln x+c$
If we use the above formula we will get,
Therefore, Integration Factor $={{e}^{\ln t}}$
To proceed further in the solution we should know the formula given below,
Formula:
${{e}^{\ln x}}=x$
Therefore we will get,
Therefore, Integration Factor = t
If we put the value of equation (3) in the above equation we will get,
Therefore, Integration Factor = log x ………………………………………… (4)
Now if we put the value of equation (4) and equation (1) in equation (2) we will get,
$\therefore y\times \log x=\int{2}\times \log x\times dx$
$\therefore y\times \log x=2\int{\log x\times dx}$
$\therefore y\times \log x=2\int{\log x\times 1\times dx}$
If we use the ILATE rule as logarithm is before algebraic therefore we will get, $\therefore y\times \log x=2\int{\log x\times 1\times dx}$ …………………………………………… (5)
Now to solve the above equation we should know the formula given below,
Formula: (Integration by parts)
$\int{u\times vdx}=u\int{vdx}-\int{\left[ \dfrac{du}{dx}\times \int{vdx} \right]dx}$
Where, u and v can be found by using ILATE rule,
As we have used ILATE rule in equation (5) therefore we can use the formula in equation (5) therefore we will get,
$\therefore y\times \log x=2\int{\log x\times 1\times dx}$
$\therefore y\times \log x=2\left\{ \log x\int{1\times dx}-\int{\left[ \dfrac{d}{dx}\left( \log x \right)\int{1\times dx} \right]dx} \right\}+C$
As we know that, $\int{1dx}=x$ and $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ therefore we can write,
$\therefore y\times \log x=2\left\{ \log x\times x-\int{\left[ \dfrac{1}{x}\times x \right]dx} \right\}+C$
$\therefore y\times \log x=2\left\{ x\log x-\int{1\times dx} \right\}+C$
$\therefore y\times \log x=2\left\{ x\log x-x \right\}+C$
$\therefore y\times \log x=2x\left( \log x-1 \right)+C$……………………………… (6)
Now to find the value of ‘C’ we will write the equation (i),
$x\log x\dfrac{dy}{dx}+y=2x\log x$
As we have given that $x\ge 1$ therefore we will use the boundary condition i.e. we will put x = 1 in the above equation therefore we will get,
$1\times \log 1\times \dfrac{dy}{dx}+y=2\times 1\times \log 1$
As log 1 = 0 therefore we will get,
$\therefore 0+y=0$
$\therefore y=0$
If we put x = 1 and y = 0 in equation (6) we will get,
$\therefore 0\times \log 1=2\times 1\times \left( \log 1-1 \right)+C$
As the value of log 1 = 0, therefore we will get,
$\therefore 0=2\left( 0-1 \right)+C$
$\therefore 0=-2+C$
$\therefore C=2$
Now put the value of ‘c’ in equation (6) therefore we will get,
$\therefore y\times \log x=2x\left( \log x-1 \right)+2$
Now, to find the value of y(e) we will put x = e in the above equation therefore we will get,
$\therefore y\times \log e=2e\left( \log e-1 \right)+2$
As the value of log e =1 therefore we will get,
$\therefore y\times 1=2e\left( 1-1 \right)+2$
$\therefore y=0+2$
$\therefore y=2$
Therefore the value of y(e) is equal to 2
Therefore the correct answer is option (c).

Note: Do remember that while writing the integration by parts students generally forgot to write the constant ‘C’ as they are not very aware of that but in this particular problem if you write the solution without using ‘C’ then your answer will definitely become wrong which is shown below,
Solution without ‘C’
$\therefore y\times \log x=2x\left( \log x-1 \right)$
$\therefore y\left( e \right)=0$ Which is a wrong answer.