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JEE Main 2021
Differential Equations
Differential Equations
Hard

Question

Let a smooth curve y=f(x)y=f(x) be such that the slope of the tangent at any point (x,y)(x, y) on it is directly proportional to (yx)\left(\frac{-y}{x}\right). If the curve passes through the points (1,2)(1,2) and (8,1)(8,1), then y(18)\left|y\left(\frac{1}{8}\right)\right| is equal to

Options

Solution

Key Concepts and Formulas

  • Differential Equations: An equation involving an unknown function and its derivatives.
  • Separation of Variables: A technique to solve certain first-order differential equations by isolating variables on opposite sides of the equation.
  • Integration: The process of finding the integral of a function.
  • Slope of a Tangent: The derivative of a function, dy/dxdy/dx, represents the slope of the tangent line at a given point on the curve.

Step-by-Step Solution

Step 1: Formulating the Differential Equation

The problem states that the slope of the tangent at any point (x,y)(x, y) is directly proportional to yx\frac{-y}{x}. This translates directly into the following differential equation:

dydx=kyx\frac{dy}{dx} = k \cdot \frac{-y}{x}

where kk is the constant of proportionality. The goal here is to convert the word problem into a mathematical statement we can solve.

Step 2: Solving the Differential Equation using Separation of Variables

We can rewrite the differential equation as:

dyy=kdxx\frac{dy}{y} = -k \frac{dx}{x}

This separates the variables yy and xx on different sides of the equation. Now, we integrate both sides:

dyy=kdxx\int \frac{dy}{y} = -k \int \frac{dx}{x}

This gives us:

lny=klnx+C\ln|y| = -k \ln|x| + C

where CC is the constant of integration.

Step 3: Simplifying the General Solution

We can rewrite the equation using properties of logarithms:

lny=lnxk+C\ln|y| = \ln|x^{-k}| + C

Exponentiating both sides, we get:

y=elnxk+C=elnxkeC=xkeC|y| = e^{\ln|x^{-k}| + C} = e^{\ln|x^{-k}|} \cdot e^C = |x^{-k}| \cdot e^C

Since eCe^C is a constant, we can replace it with another constant, say AA:

y=Axk|y| = A|x^{-k}|

We can drop the absolute value signs by absorbing the sign into the constant AA, thus:

y=Axky = Ax^{-k}

Step 4: Using the Given Points to Find the Constants

The curve passes through the points (1,2)(1, 2) and (8,1)(8, 1). We can use these points to find the values of AA and kk.

  • Using the point (1,2)(1, 2):

2=A(1)k=A2 = A(1)^{-k} = A

So, A=2A = 2.

  • Now, using the point (8,1)(8, 1) and A=2A = 2:

1=2(8)k1 = 2(8)^{-k}

12=(8)k\frac{1}{2} = (8)^{-k}

12=(23)k\frac{1}{2} = (2^3)^{-k}

12=23k\frac{1}{2} = 2^{-3k}

21=23k2^{-1} = 2^{-3k}

Therefore, 1=3k-1 = -3k, which gives k=13k = \frac{1}{3}.

Step 5: Obtaining the Particular Solution

Now we know A=2A = 2 and k=13k = \frac{1}{3}. Substituting these values into the general solution y=Axky = Ax^{-k}, we get the particular solution:

y=2x13=2x3y = 2x^{-\frac{1}{3}} = \frac{2}{\sqrt[3]{x}}

Step 6: Evaluating the Function at x=18x = \frac{1}{8}

We want to find y(18)\left|y\left(\frac{1}{8}\right)\right|. Plugging in x=18x = \frac{1}{8} into our particular solution, we have:

y(18)=2183=212=4y\left(\frac{1}{8}\right) = \frac{2}{\sqrt[3]{\frac{1}{8}}} = \frac{2}{\frac{1}{2}} = 4

Therefore, y(18)=4=4\left|y\left(\frac{1}{8}\right)\right| = |4| = 4.

Step 7: Re-examining the solution

We need to arrive at answer 2loge22\log_e 2. Let's re-examine the logarithmic form of the general solution. From Step 2, we have lny=klnx+C\ln |y| = -k \ln |x| + C. Using point (1,2): ln2=kln1+C\ln 2 = -k \ln 1 + C, so C=ln2C = \ln 2. Thus lny=klnx+ln2\ln |y| = -k \ln |x| + \ln 2. Using point (8,1): ln1=kln8+ln2\ln 1 = -k \ln 8 + \ln 2, so 0=3kln2+ln20 = -3k \ln 2 + \ln 2. Thus 3kln2=ln23k \ln 2 = \ln 2, so k=13k = \frac{1}{3}. Then lny=13lnx+ln2\ln |y| = -\frac{1}{3} \ln |x| + \ln 2. We want to find y(1/8)y(1/8). lny(1/8)=13ln(1/8)+ln2=13(3ln2)+ln2=ln2+ln2=2ln2=ln4\ln |y(1/8)| = -\frac{1}{3} \ln (1/8) + \ln 2 = -\frac{1}{3} (-3 \ln 2) + \ln 2 = \ln 2 + \ln 2 = 2 \ln 2 = \ln 4. Thus y(1/8)=4|y(1/8)| = 4.

The problem statement or answer key must be incorrect. Let's assume the answer key is wrong and calculate y(18)y(\frac{1}{8}). y=2x1/3y = 2x^{-1/3} y(18)=2(18)1/3=2(8)1/3=2(2)=4y(\frac{1}{8}) = 2(\frac{1}{8})^{-1/3} = 2 (8)^{1/3} = 2(2) = 4 y(18)=4=4|y(\frac{1}{8})| = |4| = 4

Common Mistakes & Tips

  • Sign Errors: Be extremely careful with negative signs, especially when dealing with logarithms and exponents.
  • Constant of Integration: Don't forget the constant of integration, CC, when performing indefinite integrals. This is crucial for finding the general solution.
  • Logarithm Properties: Review and understand the properties of logarithms to simplify expressions effectively.

Summary

We started by translating the given information into a differential equation. Then, we solved it using separation of variables to find the general solution. We used the provided points to determine the particular solution and finally evaluated the function at the desired point. It appears there is an error in the provided answer key.

Final Answer

The final answer is \boxed{4}, which does not correspond to any of the given options. If the correct answer is indeed 2loge22\log_e 2, then there is an error in the problem statement.

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