Textbook Question
21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.
y'(t) = 3 + e⁻²ᵗ
39
views
13. Intro to Differential Equations
Basics of Differential Equations
3:59 minutesProblem 9.1.51c
Textbook Question
Another second-order equation Consider the differential equation y''(t) + k²y(t) = 0, where k is a positive real number.
c. Give the general solution of the equation for arbitrary k > 0 and verify your conjecture.
Verified step by step guidance1
Recognize that the given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \(y''(t) + k^{2} y(t) = 0\), where \(k > 0\).
Write the characteristic equation associated with the differential equation by replacing \(y(t)\) with \(e^{\lambda t}\), which gives: \(\lambda^{2} + k^{2} = 0\).
Solve the characteristic equation for \(\lambda\): \(\lambda^{2} = -k^{2}\), so \(\lambda = \pm i k\), where \(i\) is the imaginary unit.
Since the roots are purely imaginary, the general solution to the differential equation is a linear combination of sine and cosine functions: \(y(t) = C_{1} \cos(k t) + C_{2} \sin(k t)\), where \(C_{1}\) and \(C_{2}\) are arbitrary constants.
Verify the solution by differentiating \(y(t)\) twice to find \(y''(t)\), then substitute \(y(t)\) and \(y''(t)\) back into the original equation \(y''(t) + k^{2} y(t) = 0\) to confirm that it holds true for all \(t\).
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Second-Order Linear Homogeneous Differential Equations
These are differential equations involving the second derivative of a function, with no terms independent of the function or its derivatives. The general form is y'' + p(t)y' + q(t)y = 0. Solutions depend on the characteristic equation derived from constant coefficients, which determines the behavior of the solution.
Recommended video:
07:39
Classifying Differential Equations
Characteristic Equation and Roots
For constant coefficient equations like y'' + k²y = 0, the characteristic equation is r² + k² = 0. Solving this yields complex roots ±ik, indicating oscillatory solutions. The nature of these roots guides the form of the general solution using sine and cosine functions.
Recommended video:
07:15Root Test
General Solution of Homogeneous Equations with Complex Roots
When the characteristic roots are complex conjugates α ± βi, the general solution is y(t) = e^{αt}(C₁ cos(βt) + C₂ sin(βt)). For purely imaginary roots (α=0), the solution simplifies to y(t) = C₁ cos(kt) + C₂ sin(kt), representing oscillations with frequency k.
Recommended video:
04:00Solutions to Basic Differential Equations
Related Videos
Related Practice