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 â€˜Câ€™ 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,
$x\log x\dfrac{dy}{dx}+y=2x\log x$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (i)
Now to simplify the equation we will divide it by â€˜xlogxâ€™ 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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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, $\therefore y\times \log x=2\int{\log x\times 1\times dx}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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,
$\therefore y\times \log x=2\left\{ \log x\times x-\int{\left[ \dfrac{1}{x}\times x \right]dx} \right\}+C$
$\therefore y\times \log x=2\left\{ x\log x-\int{1\times dx} \right\}+C$
$\therefore y\times \log x=2\left\{ x\log x-x \right\}+C$
$\therefore y\times \log x=2x\left( \log x-1 \right)+C$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (6)
Now to find the value of â€˜Câ€™ 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 â€˜câ€™ 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 â€˜Câ€™ as they are not very aware of that but in this particular problem if you write the solution without using â€˜Câ€™ then your answer will definitely become wrong which is shown below,
Solution without â€˜Câ€™
$\therefore y\times \log x=2x\left( \log x-1 \right)$
$\therefore y\left( e \right)=0$ Which is a wrong answer.