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Chapter 2: Problem 23

Let \(M=4 y+2 t-5\) and \(N=6 y+4 t-1\) so that \(M_{y}=4=N_{t} .\) From \(f_{t}=4 y+2 t-5\) we obtain \(f=4 t y+t^{2}-5 t+h(y), h^{\prime}(y)=6 y-1,\) and \(h(y)=3 y^{2}-y .\) The solution is \(4 t y+t^{2}-5 t+3 y^{2}-y=c .\) If \(y(-1)=2\) then \(c=8\) and a solution of the initial-value problem is \(4 t y+t^{2}-5 t+3 y^{2}-y=8\)

Short Answer

Expert verified
The particular solution is \(4ty + t^2 - 5t + 3y^2 - y = 8\).

Step by step solution

01

Analyze Given Functions M and N

We start with the given expressions for \(M\) and \(N\). \(M=4y+2t-5\) and \(N=6y+4t-1\). Identify partial derivatives: \(M_y=4\) and \(N_t=4\). These derivatives will be important for the upcoming steps.
02

Set Up the Function f

We are given the derivative \(f_t = 4y + 2t - 5\). We need to find the function \(f\) such that its partial derivative with respect to \(t\) matches: \(f = 4ty + t^2 - 5t + h(y)\).
03

Determine Additional Function h(y)

To find \(h(y)\), compute its derivative to fit the form \(h'(y) = 6y - 1\) given in the problem. Integrate to find \(h(y) = 3y^2 - y\).
04

Complete the General Solution for f

Substitute \(h(y)\) back into the expression for \(f\): \(f = 4ty + t^2 - 5t + 3y^2 - y\). This is the general solution of the problem. Let it equal to a constant: \(4ty + t^2 - 5t + 3y^2 - y = c\).
05

Use Initial Condition to Solve for c

Given that \(y(-1) = 2\), substitute \(t = -1\) and \(y = 2\) into our general solution to find \(c\). This leads to the equation \(4(-1)(2) + (-1)^2 - 5(-1) + 3(2)^2 - 2 = c\), simplifying which gives \(c = 8\).
06

Present the Particular Solution

The solution to the initial-value problem is \(4ty + t^2 - 5t + 3y^2 - y = 8\) as found by using the initial condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
An initial value problem is a type of differential equation that comes with a specific set of conditions. These conditions are often given as the values of the function at a particular point. By knowing what happens at the start (or a defined point), you can deduce the behavior of the entire system.

In our problem, the initial condition is stated as: \( y(-1) = 2 \). This means, at time \( t = -1 \), the value of the function \( y \) is 2. By using this initial condition, we can find the constant \( c \) in our general solution, making it possible to solve for the particular solution that fits these conditions. Understanding initial value problems is crucial as it helps in predicting or controlling systems modeled by differential equations.
Partial Derivatives
Partial derivatives provide a way to see how a function changes as only one of its variables is incremented, while others are held constant. It鈥檚 like zooming into one direction of movement.

In our exercise, partial derivatives help us simplify the problem. For the functions given by the exercise, such as \( M = 4y + 2t - 5 \) and \( N = 6y + 4t - 1 \), we look at \( M_y = 4 \) and \( N_t = 4 \). This indicates a kind of symmetry or relationship between these components that is essential for finding a compatible function. These derivatives are crucial in ensuring that the integration performed later aligns properly with the equations given.
Integration
Integration is the process of finding a function from its derivative, essentially the reverse of differentiation. It is a key concept in solving differential equations.

In our context, we were given the derivative \( f_t = 4y + 2t - 5 \) and needed to find the function \( f \). By integrating with respect to \( t \), we obtained \( f = 4ty + t^2 - 5t + h(y) \). However, we couldn't stop there, because another part of our equation, \( h(y) \), was still unknown.

We integrated \( h'(y) = 6y - 1 \) to get \( h(y) = 3y^2 - y \). Each step in integration brings us closer to uncovering the full expression needed to describe our system.
General Solution
A general solution to a differential equation includes arbitrary constants and provides a family of potential solutions.

For our problem, the general solution obtained was \( 4ty + t^2 - 5t + 3y^2 - y = c \). Here, \( c \) represents an unknown constant. This solution includes all possible paths the system can take, based on the given form of the differential equations.

However, by applying the initial condition, our general solution is narrowed down to the particular solution, \( 4ty + t^2 - 5t + 3y^2 - y = 8 \). This particular solution describes the unique situation that fits the initial condition \( y(-1) = 2 \). Understanding how to transition from general to particular solutions is essential in solving initial value problems, providing insight into real-world dynamic systems.

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Most popular questions from this chapter

For \(\frac{d T}{d t}-k T=-T_{m} k\) an integrating factor is \(e^{\int(-k) d t}=e^{-k t}\) so that \(\frac{d}{d t}\left[e^{-k t} T\right]=-T_{m} k e^{-k t}\) and \(T=T_{m}+c e^{k t}\) for \(-\infty< t<\infty .\) If \(T(0)=T_{0}\) then \(c=T_{0}-T_{m}\) and \(T=T_{m}+\left(T_{0}-T_{m}\right) e^{k t}\).

(a) The second derivative of \(y\) is \\[\frac{d^{2} y}{d x^{2}}=-\frac{d y / d x}{(y-1)^{2}}=-\frac{1 /(y-3)}{(y-3)^{2}}=-\frac{1}{(y-3)^{3}}\\] The solution curve is concave down when \(d^{2} y / d x^{2}<0\) or \(y>3\), and concave up when \(d^{2} y / d x^{2}>0\) or \(y<3 .\) From the phase portrait we see that the solution curve is decreasing when \(y<3\) and increasing when \(y>3\). (b) Separating variables and integrating we obtain \\[ \begin{aligned} (y-3) d y &=d x \\ \frac{1}{2} y^{2}-3 y &=x+c \\ y^{2}-6 y+9 &=2 x+c_{1} \\ (y-3)^{2} &=2 x+c_{1} \\ y &=3 \pm \sqrt{2 x+c_{1}}.\end{aligned}\\] The initial condition dictates whether to use the plus or minus sign. When \(y_{1}(0)=4\) we have \(c_{1}=1\) and \(y_{1}(x)=3+\sqrt{2 x+1}\). When \(y_{2}(0)=2\) we have \(c_{1}=1\) and \(y_{2}(x)=3-\sqrt{2 x+1}.\) When \(y_{3}(1)=2\) we have \(c_{1}=-1\) and \(y_{3}(x)=3-\sqrt{2 x-1}.\) When \(y_{4}(-1)=4\) we have \(c_{1}=3\) and \(y_{4}(x)=3+\sqrt{2 x+3}\).

(a) An implicit solution of the differential equation \((2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0\) is \\[y^{2}+2 y-x^{4}-3 x^{2}+c=0\\]. The condition \(y(0)=-3\) implies that \(c=-3 .\) Therefore \(y^{2}+2 y-x^{4}-3 x^{2}-3=0\). (b) Using the quadratic formula we can solve for \(y\) in terms of \(x\): \\[y=\frac{-2 \pm \sqrt{4+4\left(x^{4}+3 x^{2}+3\right)}}{2}\\]. The explicit solution that satisfies the initial condition is then \\[y=-1-\sqrt{x^{4}+3 x^{3}+4}\\]. (c) From the graph of the function \(f(x)=x^{4}+3 x^{3}+4\) below we see that \(f(x) \leq 0\) on the approximate interval \(-2.8 \leq x \leq-1.3 .\) Thus the approximate domain of the function $$y=-1-\sqrt{x^{4}+3 x^{3}+4}=-1-\sqrt{f(x)}$$ is \(x \leq-2.8\) or \(x \geq-1.3 .\) The graph of this function is shown below. (d) Using the root finding capabilities of a CAS, the zeros of \(f\) are found to be -2.82202 and \(-1.3409 .\) The domain of definition of the solution \(y(x)\) is then \(x>-1.3409 .\) The equality has been removed since the derivative \(d y / d x\) does not exist at the points where \(f(x)=0 .\) The graph of the solution \(y=\phi(x)\) is given on the right.

For \(y^{\prime}+\frac{1}{x} y=\frac{1}{x} e^{x}\) an integrating factor is \(e^{\int(1 / x) d x}=x\) so that \(\frac{d}{d x}[x y]=e^{x}\) and \(y=\frac{1}{x} e^{x}+\frac{c}{x}\) for \(0< x<\infty\). If \(y(1)=2\) then \(c=2-e\) and \(y=\frac{1}{x} e^{x}+\frac{2-e}{x}\).

Separating variables and integrating we obtain \\[\frac{d x}{\sqrt{1-x^{2}}}-\frac{d y}{\sqrt{1-y^{2}}}=0 \quad \text { and } \quad \sin ^{-1} x-\sin ^{-1} y=c\\]. Setting \(x=0\) and \(y=\sqrt{3} / 2\) we obtain \(c=-\pi / 3 .\) Thus, an implicit solution of the initial-value problem is \(\sin ^{-1} x-\sin ^{-1} y=\pi / 3 .\) Solving for \(y\) and using an addition formula from trigonometry, we get \\[y=\sin \left(\sin ^{-1} x+\frac{\pi}{3}\right)=x \cos \frac{\pi}{3}+\sqrt{1-x^{2}} \sin \frac{\pi}{3}=\frac{x}{2}+\frac{\sqrt{3} \sqrt{1-x^{2}}}{2}\\].
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