Question

If y(x) is the solution of the differential equation \[ (1+x^2)y'' - 2xy' = 0 \] satisfying y(0) = 0 and y'(0) = 3, then y(1) equals ....................

Hint:

Recognize ODE forms that allow for simplification. A second-order ODE of the form \(f(x, y', y'')=0\), where the dependent variable \(y\) is missing, can always be reduced to a first-order ODE by the substitution \(v=y'\).

Answer 1

Step 1: Understanding the Concept:
The problem is an initial value problem for a second-order linear homogeneous ordinary differential equation. The equation can be solved using the method of reduction of order.
Step 2: Key Formula or Approach:
1. Let \(v = y'\). This transforms the second-order ODE for \(y\) into a first-order ODE for \(v\). 2. The new equation is \(v' = y''\), so \((1+x^2)v' - 2xv = 0\). 3. Solve this first-order separable equation for \(v\). 4. Substitute back \(v = y'\) and integrate to find \(y\). 5. Use the initial conditions \(y(0)=0\) and \(y'(0)=3\) to determine the constants of integration. 6. Evaluate the final solution at \(x=1\).
Step 3: Detailed Explanation or Calculation:
Given the differential equation: \[ (1+x^2)y'' - 2xy' = 0 \] Let \(v(x) = y'(x)\). Then \(v'(x) = y''(x)\). Substituting these into the equation gives: \[ (1+x^2)v' - 2xv = 0 \] This is a first-order linear and separable differential equation. We can write it as: \[ (1+x^2)\frac{dv}{dx} = 2xv \] Separate the variables \(v\) and \(x\): \[ \frac{dv}{v} = \frac{2x}{1+x^2} dx \] Integrate both sides: \[ \int \frac{1}{v} dv = \int \frac{2x}{1+x^2} dx \] \[ \ln|v| = \ln(1+x^2) + C_1 \] Exponentiate both sides to solve for \(v\): \[ v = e^{\ln(1+x^2) + C_1} = e^{C_1}e^{\ln(1+x^2)} = C(1+x^2) \] where \(C = e^{C_1}\) is an arbitrary constant.
Now, substitute back \(v = y'\): \[ y'(x) = C(1+x^2) \] Use the first initial condition, \(y'(0) = 3\): \[ 3 = C(1 + 0^2) \implies C = 3 \] So, the expression for the first derivative is: \[ y'(x) = 3(1+x^2) \] To find \(y(x)\), integrate \(y'(x)\) with respect to \(x\): \[ y(x) = \int 3(1+x^2) dx = 3 \left(x + \frac{x^3}{3}\right) + D = 3x + x^3 + D \] Use the second initial condition, \(y(0) = 0\): \[ 0 = 3(0) + (0)^3 + D \implies D = 0 \] The particular solution is: \[ y(x) = x^3 + 3x \] Finally, evaluate \(y(x)\) at \(x=1\): \[ y(1) = (1)^3 + 3(1) = 1 + 3 = 4 \] Step 4: Final Answer:
The value of y(1) is 4.
Step 5: Why This is Correct:
The step-by-step solution by reduction of order correctly solves the differential equation. The constants of integration are determined using the given initial conditions, leading to the unique solution \(y(x)=x^3+3x\). Evaluating this at \(x=1\) gives the value 4.