Let \( y = y(x) \) be the solution of the differential equation \[\left( 2x \log_e x \right) \frac{dy}{dx} + 2y = \frac{3}{x} \log_e x, \, x>0 \, \text{and} \, y(e^{-1}) = 0.\] Then, \( y(e) \) is equal to:
- \( \frac{3}{2e} \)
- \( \frac{-2}{3e} \)
- \( \frac{-3}{e} \)
- \( \frac{-2}{e} \)
The Correct Option is C
Approach Solution - 1
The given differential equation is:
\[ \frac{dy}{dx} + \frac{y}{x \ln x} = \frac{3}{2x^2}. \]
Step 1: Find the integrating factor (I.F.)
\[ \text{I.F.} = e^{\int \frac{1}{x \ln x} dx} = e^{\ln(\ln x)} = \ln x. \]
Step 2: Multiply through by I.F.
\[ (\ln x)y = \int \frac{3 \ln x}{2x^2} dx. \]
Step 3: Solve the integral
\[ \int \frac{3 \ln x}{2x^2} dx = \frac{3}{2} \int x^{-2} \ln x dx. \]
Use integration by parts, letting \( u = \ln x \) and \( dv = x^{-2} dx \):
\[ \int \ln x \cdot x^{-2} dx = -\frac{\ln x}{x} - \int -\frac{1}{x^2} dx = -\frac{\ln x}{x} + \frac{1}{x}. \]
Thus:
\[ \int \frac{3 \ln x}{2x^2} dx = \frac{3}{2} \left( -\frac{\ln x}{x} + \frac{1}{x} \right). \]
Step 4: Write the solution
\[ y \ln x = \frac{3}{2} \left( -\frac{\ln x}{x} + \frac{1}{x} \right) + C. \]
Simplify:
\[ y = -\frac{3 \ln x}{2x \ln x} + \frac{3}{2x \ln x} + \frac{C}{\ln x}. \]
\[ y = -\frac{3}{2x} + \frac{C}{\ln x}. \]
Step 5: Apply the initial condition \( y(e^{-1}) = 0 \)
\[ 0 = -\frac{3}{2e^{-1}} + \frac{C}{\ln(e^{-1})}. \]
\[ 0 = -\frac{3}{2e} + \frac{C}{-1}. \]
\[ C = -\frac{3}{2}e. \]
Step 6: Find \( y(e) \)
\[ y(e) = -\frac{3}{2e} + \frac{-\frac{3}{2}e}{\ln e}. \]
\[ y(e) = -\frac{3}{2e} - \frac{3}{2e} = -\frac{3}{e}. \]
Final Answer: \(-\frac{3}{e}\).
Approach Solution -2
To solve the given differential equation, we need to find the value of \( y \) at \( x = e \). Given the differential equation:
\(\left( 2x \log_e x \right) \frac{dy}{dx} + 2y = \frac{3}{x} \log_e x\)
We can rearrange it to form a standard linear differential equation in the form of:
\(\frac{dy}{dx} + P(x)y = Q(x)\)
Where:
- \(P(x) = \frac{2}{2x \log_e x} = \frac{1}{x \log_e x}\)
- \(Q(x) = \frac{3}{2x^2 (\log_e x)^2}\)
The integrating factor (IF) is given by:
\(\text{IF} = e^{\int P(x) \, dx} = e^{\int \frac{1}{x \log_e x} \, dx}\)
Let \(u = \log_e x\), then \(\frac{du}{dx} = \frac{1}{x}\), so \(dx = x \, du\).
Thus, the integral becomes:
\(\int \frac{1}{x \log_e x} \, dx = \int \frac{1}{u} \, du = \log_e |u| + C = \log_e |\log_e x| + C\)
Therefore, the integrating factor is:
\(\text{IF} = e^{\log_e |\log_e x|} = |\log_e x|\)
We multiply the entire differential equation by the integrating factor:
\(|\log_e x|\left( \frac{dy}{dx} + \frac{y}{x \log_e x} \right) = \frac{3}{2x^2 (\log_e x)}\)
It simplifies to:
\(\frac{d}{dx}(y|\log_e x|) = \frac{3}{2x^2}\)
Integrating both sides with respect to \(x\):
\(y|\log_e x| = \int \frac{3}{2x^2} \, dx = -\frac{3}{2x} + C\)
Solving for \(y\), we get:
\(y = \frac{-3}{2x|\log_e x|} + \frac{C}{|\log_e x|}\)
Using the initial condition \(y(e^{-1}) = 0\):
\(0 = \frac{-3}{2e^{-1}|\log_e e^{-1}|} + \frac{C}{|\log_e e^{-1}|}\)
Simplifying:
\(0 = \frac{-3}{2\cdot e^{-1} \cdot 1} + \frac{C}{1}\)
The solving for \(C\) gives: \(C = \frac{3}{2e^{-1}}\)
Finally, finding \( y \) at \( x = e \):
\(y(e) = \frac{-3}{2e \cdot 1} + \frac{\frac{3}{2}e}{1} = \frac{-3}{2e} + \frac{3}{2e} = \frac{-3}{e}\)
Thus, the value of \( y(e) \) is \(\frac{-3}{e}\),
Therefore, the correct answer is: \(\frac{-3}{e}\)
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Differential equations Questions
Let $y=y(x)$ be the solution of the differential equation $\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1$.Then $6 y^2( e )$ is equal to
- If \((t+1)dx=(2x+(t+1)^3)dt\) and \(x(0)=2\), then \(x(1) \)is equal to:
- Let vertex \(A(2, 3, 1), B(3, 2, -1), C(-2, 1, 3)\). If \(AD\) is angle bisector of angle \(A\), then projection of \(\overrightarrow{AD}\) on \(\overrightarrow{AC}\) is equal to:
- Let \(y = y(x)\) be the solution of the differential equation \(\sec(x) \frac{dy}{dx} + [2(1 - x) \tan(x) + x(2 - x)] = 0\) such that \(y(0) = 2\). Then \(y(2)\) is equal to:
- Let \( y = y(x) \) be the solution of the differential equation \( (1 - x^2) \, dy = \left[ xy + \left( x^3 + 2 \right) \sqrt{3 \left( 1 - x^2 \right)} \right] dx \), \( -1 < x < 1, y(0) = 0 \). If \( y\left( \frac{1}{2} \right) = \frac{m}{n} \), \( m \) and \( n \) are coprime numbers, then \( m + n \) is equal to \(\_\_\_\_\_\_\_\_\_\).
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,