Question

The function \(y(t)\) is a solution of the initial value problem \(y^{\prime \prime}+a y^{\prime}+b y=0, y\left(t_{0}\right)=y_{0}\), \(y^{\prime}\left(t_{0}\right)=y_{0}^{\prime}\), where the point \(t_{0}\) is specified. Determine the constants \(a, b, y_{0}\), and \(y_{0}^{\prime} .\) $$ y(t)=\sin t-\sqrt{2} \cos t, \quad t_{0}=\pi / 4 $$

Short answer

Based on the given function \(y(t) = \sin t - \sqrt{2}\cos t\) and the initial point \(t_0 = \frac{\pi}{4}\), we have found the following constants for the associated differential equation \(y'' + ay' + by = 0\): Constant a: \(a = -1\) Constant b: \(b = \sqrt{2}\) Initial value y_0: \(y_0 = 1 - \sqrt{2}\) Initial value of y': \(y_0' = 1+\sqrt{2}\)

Step-by-step solution

Compute y'(t) and y''(t)

To find the first and second derivatives of y with respect to t, differentiate \(y(t)\) with respect to \(t\) twice: First derivative, \(y'(t)\): $$ y'(t)=\frac{d}{dt}(\sin t-\sqrt{2}\cos t) = \cos t + \sqrt{2}\sin t $$ Second derivative, \(y''(t)\): $$ y''(t)=\frac{d^2}{dt^2}(\sin t - \sqrt{2}\cos t) = -\sin t + 2\cos t $$

Compute y(t_0) and y'(t_0)

To find the values of \(y_0\) and \(y_0'\), for the given \(t_0 = \frac{\pi}{4}\), substitute \(t_0\) into the expressions of \(y(t)\) and \(y'(t)\): Find y_0: $$ y_0= y\left(\frac{\pi}{4}\right)=\sin{\frac{\pi}{4}}-\sqrt{2}\cos{\frac{\pi}{4}} =\frac{1}{\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{2}} = 1-\sqrt{2} $$ Find y_0': $$ y_{0}'= y'\left(\frac{\pi}{4}\right)=\cos{\frac{\pi}{4}}+\sqrt{2}\sin{\frac{\pi}{4}} =\frac{1}{\sqrt{2}}+\frac{1\sqrt{2}}{\sqrt{2}} = 1+\sqrt{2} $$

Find values of a and b using the differential equation

Substitute \(y(t)\), \(y'(t)\), and \(y''(t)\) in the given differential equation \(y''+ay'+by=0\) : $$ (-\sin t + 2\cos t) + a(\cos t + \sqrt{2}\sin t) + b(\sin t - \sqrt{2}\cos t) = 0 $$ Now, substitute \(t_0\), \(y_0\), and \(y_0'\) from Step 2 into the differential equation: $$ (-\sin{\frac{\pi}{4}} + 2\cos{\frac{\pi}{4}})+a(\cos{\frac{\pi}{4}}+\sqrt{2}\sin{\frac{\pi}{4}})+b(\sin{\frac{\pi}{4}}-\sqrt{2}\cos{\frac{\pi}{4}}) = 0 $$ Simplify the equation: $$ (-\frac{1}{\sqrt{2}} + \sqrt{2})+a(\frac{1}{\sqrt{2}}+\sqrt{2})+b(\frac{1}{\sqrt{2}}-\sqrt{2}) = 0 $$ Substitute the values of \(y_0\) and \(y_0'\): $$ (1-\sqrt{2})+a(1+\sqrt{2})+b(1-\sqrt{2}) = 0 $$ Now we can easily solve it for the constants a and b. The equation simplifies to: $$ (1 - \sqrt{2}) - a\sqrt{2} + a + b - b\sqrt{2} = 0 $$ Since \(y(t)\) is a solution to the given differential equation, coefficients of sin t and cos t terms must vanish: Coefficient of sin t (a√2 + b) = \(0\). Then we get \(a\sqrt{2}+b=0 \Rightarrow a\sqrt{2}=-b\). This is our first equation. Coefficient of cos t (1 + a - b√2) = \(0\). Then we get \(1 + a - b\sqrt{2} =0 \Rightarrow a-b\sqrt{2}= -1\). This is our second equation. Now we can solve these two equations using substitution or elimination method. Let's use substitution. From the first equation, we find \(b = -a\sqrt{2}\). Substituting this value for \(b\) in the second equation, we get: $$ a - (-a\sqrt{2})\sqrt{2} = -1 $$ Solving for \(a\), we find that \(a = -1\). Then, substituting \(a = -1\) in the first equation, we get \(b = \sqrt{2}\).

Summary

The constants for the given initial value problem are: $$ a = -1\\ b = \sqrt{2}\\ y_0 = 1 - \sqrt{2}\\ y_0' = 1+\sqrt{2} $$

Key concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They play a crucial role in mathematics, physics, engineering, biology, and many other fields because they describe the rate at which changes occur. The general form of a nth order differential equation is

\( F(t, y, y', y'', ... , y^{(n)}) = 0 \),

where F is a given function, t is the independent variable (often representing time), and y is the unknown function we are trying to find, with y', y'', ..., y^{(n)} being its first, second, ..., nth derivatives, respectively. Differential equations can be either ordinary (ODEs), involving one independent variable, or partial (PDEs), involving multiple independent variables. In the exercise provided, we are dealing with a second-order ODE, which has the form \( y'' + ay' + by = 0 \),
where a and b are constants to be determined.

Derivatives

Derivatives represent the rate at which a function changes at any given point, and they are a fundamental tool in calculus. When we take the derivative of a function, we are finding the slope of the tangent to the function at any point. Mathematically, the derivative of a function y(t) with respect to t is denoted as y'(t) or \( \frac{dy}{dt} \).
In the initial value problem exercise, calculating the derivatives y'(t) and y''(t) allows us to find the rate of change and the acceleration of the function y(t), respectively. These calculations are essential steps in solving the problem because they give us the necessary information to match the terms of the differential equation and find the values of the constants a and b.

Solving Initial Value Problems

Solving an initial value problem (IVP) means finding a function y(t) that not only satisfies the differential equation, but also meets the specific initial conditions given. Initial conditions typically specify values of the function and possibly one or more derivatives at a particular point t_{0}. For the example at hand,
we are given the initial conditions \( y(t_{0}) = y_{0} \) and \( y'(t_{0}) = y_{0}' \),
which fix the solution uniquely. In practice, solving an IVP often involves integrating the differential equation and using the initial conditions to determine the arbitrary constants from the general solution. In this case, initial conditions have guided us in finding the specific solution to the second-order ODE, outlining the unique trajectory based on how the function behaves at \( t_{0} = \frac{\pi}{4} \).
The solution process included calculation of derivatives, substitution of the initial values into these derivatives, and finally, resolving the coefficients in the differential equation to determine the constants. By following these steps, we can confidently determine the values of a, b, y_{0}, and y_{0}' that fulfill both the differential equation and the initial conditions provided in the exercise.