Learn how to use an initial condition to find a specific constant of integration (C), transforming a general solution into a unique particular solution for an initial value problem (IVP).
An initial value problem consists of a differential equation paired with an initial condition, such as \(\frac{dy}{dx} = f(x,y)\) with \(y(x_0) = y_0\). The goal is to find the specific solution that satisfies both.
What is the procedure for solving an initial value problem?
1. Find the general solution of the differential equation, which will include a constant \(C\).
2. Substitute the values from the initial condition (\(x_0\) and \(y_0\)) into the general solution.
3. Solve for the constant \(C\).
4. Write the particular solution with the calculated value of \(C\).
Solve the IVP: \(\frac{dy}{dx} = 2x\), with \(y(1) = 5\).
Solve the IVP: \(\frac{dy}{dx} = \frac{1}{y}\), with \(y(0) = 3\).
1. Separate and integrate: \(\int y \, dy = \int dx \implies \frac{1}{2}y^2 = x + C\).
2. Substitute \(x=0\) and \(y=3\): \(\frac{1}{2}(3)^2 = 0 + C \implies C = \frac{9}{2}\).
3. Particular solution (implicit): \(\frac{1}{2}y^2 = x + \frac{9}{2}\) or \(y^2 = 2x+9\).
From the implicit solution \(y^2 = 2x+9\) with initial condition \(y(0)=3\), find the explicit particular solution.
The explicit form is \(y = \pm \sqrt{2x+9}\). Since the initial value for \(y\) is positive (\(y=3\)), we must choose the positive square root. The explicit particular solution is \(y = \sqrt{2x+9}\).
What does a particular solution represent geometrically?
The general solution represents an infinite family of curves. A particular solution, determined by an initial condition, represents the single, unique curve from that family that passes through the specified initial point \((x_0, y_0)\).
Solve the IVP: \(\frac{dy}{dt} = -0.5y\), with \(y(0) = 100\).
The general solution to \(\frac{dy}{dt}=ky\) is \(y=Ce^{kt}\). Here, \(k=-0.5\), so \(y = Ce^{-0.5t}\). Substitute \(t=0\) and \(y=100\): \(100 = Ce^0 \implies C=100\). The particular solution is \(y = 100e^{-0.5t}\).
Find the particular solution to \(y' = y\cos(x)\) satisfying \(y(0) = 2\).
