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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
9장, 문제 9.R.33b

A second-order equation Consider the equation


t² y′′(t) + 2ty′(t) − 12y(t) = 0


b. Assuming the general solution of the equation is


y(t) = C₁ tᵖ¹ + C₂ tᵖ²,


find the solution that satisfies the conditions


y(1) = 0, y′(1) = 7

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1
Rewrite the given differential equation: \(t^{2} y''(t) + 2t y'(t) - 12 y(t) = 0\). This is a Cauchy-Euler (or equidimensional) equation, which suggests trying a solution of the form \(y(t) = t^{p}\).
Substitute \(y(t) = t^{p}\) into the differential equation. Compute the derivatives: \(y'(t) = p t^{p-1}\) and \(y''(t) = p (p-1) t^{p-2}\). Substitute these into the equation to get an algebraic equation in terms of \(p\).
After substitution, the equation becomes \(t^{2} imes p (p-1) t^{p-2} + 2t imes p t^{p-1} - 12 t^{p} = 0\). Simplify powers of \(t\) to get \(p (p-1) t^{p} + 2 p t^{p} - 12 t^{p} = 0\), which simplifies to \((p^{2} + p - 12) t^{p} = 0\).
Since \(t^{p} eq 0\) for \(t > 0\), set the characteristic equation \(p^{2} + p - 12 = 0\). Solve this quadratic equation to find the roots \(p_1\) and \(p_2\).
Write the general solution as \(y(t) = C_1 t^{p_1} + C_2 t^{p_2}\). Use the initial conditions \(y(1) = 0\) and \(y'(1) = 7\) to form a system of equations in \(C_1\) and \(C_2\). Compute \(y'(t)\), substitute \(t=1\), and solve the system to find \(C_1\) and \(C_2\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Second-Order Linear Differential Equations

These are differential equations involving the second derivative of a function. The given equation is linear and homogeneous, meaning it can be solved by finding characteristic solutions that satisfy the equation. Understanding the structure helps in applying appropriate solution methods.
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For equations of the form t²y'' + 2ty' - 12y = 0, solutions often take the form y = t^p. Substituting this form leads to a characteristic equation in terms of p, whose roots determine the general solution. This method transforms the differential equation into an algebraic problem.
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Initial Conditions and Solving for Constants

Initial conditions like y(1) = 0 and y'(1) = 7 allow us to find the specific constants C₁ and C₂ in the general solution. By substituting t = 1 into y(t) and y'(t), we create a system of equations to solve for these constants, yielding the particular solution.
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Initial Value Problems