Question

Let y=y(x) be the solution of the differential equation $x\dfrac{{dy}}{{dx}} + y = x{\log _e}x,(x > 1)$. If $2y(2) = {\log _e}4 - 1$, then $y(e)$is equal to:
A $\dfrac{{{e^2}}}{4}$
B $\dfrac{e}{4}$
C $- \dfrac{e}{2}$
D $- \dfrac{{{e^2}}}{2}$

Answer 1

In this question we have been given a differential equation $x\dfrac{{dy}}{{dx}} + y = x{\log _e}x,(x > 1)$, whose solution is y(x). So, we need to find the value of y(e). For that firstly we will divide this equation by x, so as to convert it in the form of $\dfrac{{dy}}{{dx}} + Py = Q$. After that we will be calculating the integrating factor, $IF = {e^{\int {\left( P \right)dx} }}$ which we will use to find the value of $y.x$. We will use integration by parts and ILATE rule for evaluating the integral. Then we will put the values of x from $2y(2) = {\log _e}4 - 1$ and find the value of y(e).

Complete step by step answer:
We have been provided with a differential equation: $x\dfrac{{dy}}{{dx}} + y = x{\log _e}x,(x > 1)$,
Now we will be dividing the whole equation by x: $\dfrac{{dy}}{{dx}} + y\left( {\dfrac{1}{x}} \right) = {\log _e}x,$
So, our equation is in the form: $\dfrac{{dy}}{{dx}} + Py = Q$
Now we will be finding the integrating factor using the formula: $IF = {e^{\int {\left( P \right)dx} }}$,
Keeping the values, we get: $IF = {e^{\int {\left( {\dfrac{1}{x}} \right)dx} }}$,
Now integrating using the formula: $\int {\dfrac{1}{x}dx = \log x}$,
The integrating factor would be: $IF = {e^{{{\log }_e}x}}$,
Which will be equal to: $IF = x$,
Hence, the solution of the differential equation is given by: $y.x = \int {x{{\log }_e}xdx + c}$,
So, we will be using the integration by parts method whose formula is: $\int {uvdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } } dx$.
Here we have to use the ILATE rule, which will help us decide the function to be chosen as u and v. So, the preference for choosing first function, i.e u is in the order - Inverse, Logarithm, Algebra, Trigonometry and Exponential. Here, we have algebraic function and logarithmic function, so we will choose logarithmic function as u and then apply the formula as below.
So, now we will be putting the values: $\int {x{{\log }_e}xdx = {{\log }_e}x\int {xdx - \int {\left( {\dfrac{d}{{dx}}\left( {{{\log }_e}x} \right)\int {xdx} } \right)} } } dx$,
Now we will integrate using the formula: $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$,
The value of $y.x$ comes out to be: $\int {x{{\log }_e}xdx = {{\log }_e}x\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\left( {\dfrac{1}{x}\left( {\dfrac{{{x^2}}}{2}} \right)} \right)dx} } + c$,
Simplifying it further: $y.x = \dfrac{{{x^2}}}{2}{\log _e}x - \int {\dfrac{x}{2}dx} + c$,
Now we need to integrate it again: $y.x = \dfrac{{{x^2}}}{2}{\log _e}x - \dfrac{{{x^2}}}{4} + c$,
Now put x=2 as we have been given $2y(2) = {\log _e}4 - 1 + c$,
Putting the values as per, $2y(2) = {\log _e}4 - 1 = {\log _e}4 - 1 + c$,
From this we can say that c=0,
We have: $y = \dfrac{x}{2}{\log _e}x - \dfrac{1}{4}x$,
Now we will be finding the value of y(e): $y(e) = \dfrac{e}{2}{\log _e}e - \dfrac{e}{4}$,
Simplifying it further, we will get: $y(e) = \dfrac{e}{2} - \dfrac{e}{4} = \dfrac{e}{4}$.

So, the correct answer is “Option B”.

Note: In this question, while integrating don’t use the formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ when you need to integrate $\dfrac{1}{x}$, as this formula is applicable only for the cases when n > 0 and in this case n=-1, which will lead to infinity and then you won’t be able to solve it further.