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Calculus Level 5

Given that $y = f_1(x) = x$ and $y = f_2(x)$ are linearly independent solutions of the differential equation

$(1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0,$

find $f_2(x)$ , which satisfies $f_2 \left( \dfrac{1}{2} \right) = \ln 3 - 4$ .

If $f_2 \left( \dfrac{3}{4} \right) = \dfrac{a \ln b}{c} - d$ , enter your answer as $a+b+c+d$ .

Note: $a, b, c$ and $d$ are positive integers, $\gcd(a,b) = 1$ , and $b \neq 1$ is squarefree.

This is not an original problem. (Adapted from 'Differential Equations: Linear, Nonlinear, Ordinary, Partial' by A C King, J Billingham, and S R Otto.)

The answer is 16.

1 solution

Priyamvad Tripathi
Aug 7, 2016

The above problem can be easily solved with the knowledge of Abel's Theorem and wronskian operator. Abel’s Theorem: If y1 and y2 are any two solutions of the equation y″ + p(t) y′ + q(t) y = 0, where p and q are continuous on an open interval I. Then the Wronskian W(y1, y2)(t) is given by W( y1, y2 )(t) = C e^(−∫ p(t) dt) , where C is a constant that depends on y1 and y2, but not on t. The Wronskian operator is defined as W(y1, y2)(t) = y1 y′2 − y′1 y2.

Here y1=x and p(t)=(-2x/(1-x^2)) This forms a first order linear differential equation in y2. Taking into account the initial conditions given y2 comes to be y2= 2x*ln(1+x/1-x) - 4 This gives the answer as 16.

Anarchy

This is not an original problem. (Adapted from 'Differential Equations: Linear, Nonlinear, Ordinary, Partial' by A C King, J Billingham, and S R Otto.)

The answer is 16.

1 solution

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