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Chapter 2: Problem 25

Let \(M=y^{2} \cos x-3 x^{2} y-2 x\) and \(N=2 y \sin x-x^{3}+\ln y\) so that \(M_{y}=2 y \cos x-3 x^{2}=N_{x} .\) From \(f_{x}=y^{2} \cos x-3 x^{2} y-2 x\) we obtain \(f=y^{2} \sin x-x^{3} y-x^{2}+h(y), h^{\prime}(y)=\ln y,\) and \(h(y)=y \ln y-y .\) The solution is \(y^{2} \sin x-x^{3} y-x^{2}+y \ln y-y=c .\) If \(y(0)=e\) then \(c=0\) and a solution of the initial-value problem is \(y^{2} \sin x-x^{3} y-x^{2}+y \ln y-y=0\)

Short Answer

Expert verified
The solution is \(y^2 \sin x - x^3 y - x^2 + y \ln y - y = 0\).

Step by step solution

01

Calculate Partial Derivatives

To solve the problem, we need the partial derivatives of the given functions \(M\) and \(N\). The partial derivative \(M_y\) is given by \(M_y = 2y \cos x - 3x^2\), and \(N_x = 2y \sin x - 3x^2\). Given: \(M_y = N_x\).
02

Integrate with respect to x

Given \(f_x = y^2 \cos x - 3x^2 y - 2x\), we integrate it with respect to \(x\). The integral gives us: \(f = y^2 \sin x - x^3y - x^2 + h(y)\).
03

Determine Function h(y)

We find the function \(h(y)\) by considering \(h'(y)\), which has been given as \(\ln y\). The antiderivative of \(\ln y\) is computed as \(h(y) = y \ln y - y + C\), where \(C\) is the constant of integration.
04

Formulate the General Solution

Combining the results from the steps, the general solution is \(f(x, y) = y^2 \sin x - x^3 y - x^2 + y \ln y - y = c\).
05

Apply Initial Conditions

To find the constant \(c\), use the initial condition \(y(0) = e\). Evaluate \(f(0, e)\) which gives \(e^2 \sin 0 - 0^3 e - 0^2 + e \ln e - e = 0\). This simplifies to \(0 = 0\); hence, \(c = 0\).
06

Write the Particular Solution

With \(c = 0\), the particular solution of the initial value problem is \(y^2 \sin x - x^3 y - x^2 + y \ln y - y = 0\).

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in differential equations, especially when dealing with functions of several variables. They measure how a function changes as one particular variable changes, keeping other variables constant. In our problem, we considered the functions \(M\) and \(N\). To ensure that these functions have a potential function, we need their partial derivatives to be equal in the respective parts.
  • For function \(M\), we found the partial derivative \(M_y = 2y \cos x - 3x^2\).
  • For function \(N\), we found the partial derivative \(N_x = 2y \sin x - 3x^2\).
The condition \(M_y = N_x\) indicates that the function has an underlying potential, suggesting predictability in the function's behavior across different variables.
Initial Value Problem
An initial value problem (IVP) involves finding a function that satisfies a differential equation and meets certain specified conditions. In this exercise, we are given an initial condition: \(y(0) = e\). This condition provides a starting point from which we solve the differential equation.
  • We use initial conditions to find particular solutions by plugging in the values of the functions at a specific point.
  • These conditions help in determining constants from the general solution, giving us a specific, applicable solution.
With our condition, we determine that the constant of integration \(c = 0\), leading us to the particular solution for the problem.
Integration
Integration is a core process in calculus used to find functions when given their derivatives. In our exercise, we perform integration to find the potential function \(f\) that satisfies the given differential conditions.We are provided with \(f_x = y^2 \cos x - 3x^2 y - 2x\), and we integrate this with respect to \(x\) to find:\[f = y^2 \sin x - x^3y - x^2 + h(y)\]This integration step effectively reverses differentiation, allowing us to piece back to the broader function from its rate of change. Integration is crucial in reconstructing expressions that illustrate the relationship between variables over an interval.
Function of Several Variables
In calculus, a function of several variables is an expression where more than one input is involved. These functions depend on multiple parameters, such as \(x\) and \(y\), used in our exercise.A function such as \(f(x, y)\) can describe surfaces or multidimensional behavior, which is critical in fields dealing with real-world applications like physics and engineering. In this exercise:
  • Our differential involves the variables \(x\) and \(y\), highlighting the multi-variable nature of the problem.
  • We calculate partial derivatives and integrations accordingly, causing changes in one variable to be tracked while keeping others constant.
Understanding functions with several variables expands our capacity to analyze complex phenomena.
Integration with respect to a variable
Integrating with respect to a specific variable means treating that variable as dynamic while considering others as constant. This exercise demonstrates integrating with respect to \(x\) while \(y\) is treated as a constant.The step-wise integration from \(f_x\) into the potential function \(f(x, y)\) required holding \(y\) constant:An illustration of the resulting function is:\[f = y^2 \sin x - x^3y - x^2 + h(y)\]
  • The function \(h(y)\) is resolved separately because \(x\)'s integration does not cover it.
  • This process allows easy manipulation of equations in multi-variable scenarios, aiding in uncovering nuances in relationships across different dimensions.

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Most popular questions from this chapter

From \(y^{\prime}+y=y^{-1 / 2}\) and \(w=y^{3 / 2}\) we obtain \(\frac{d w}{d x}+\frac{3}{2} w=\frac{3}{2} .\) An integrating factor is \(e^{3 x / 2}\) so that \(e^{3 x / 2} w=\) \(e^{3 x / 2}+c\) or \(y^{3 / 2}=1+c e^{-3 x / 2} .\) If \(y(0)=4\) then \(c=7\) and \(y^{3 / 2}=1+7 e^{-3 x / 2}\).

From \(d A / d t=10-A / 100\) we obtain \(A=1000+c e^{-t / 100} .\) If \(A(0)=0\) then \(c=-1000\) and \(A(t)=\) \(1000-1000 e^{-t / 100}\).

(a) \(\operatorname{From} d P / d t=\left(k_{1}-k_{2}\right) P\) we obtain \(P=P_{0} e^{\left(k_{1}-k_{2}\right) t}\) where \(P_{0}=P(0)\). (b) If \(k_{1} >k_{2}\) then \(P \rightarrow \infty\) as \(t \rightarrow \infty\). If \(k_{1}=k_{2}\) then \(P=P_{0}\) for every \(t .\) If \(k_{1}< k_{2}\) then \(P \rightarrow 0\) as \(t \rightarrow \infty\).

Separating variables we obtain \(\frac{d y}{(y-1)^{2}+0.01}=d x .\) Then \\[10 \tan ^{-1} 10(y-1)=x+c \text { and } y=1+\frac{1}{10} \tan \frac{x+c}{10}.\\] Setting \(x=0\) and \(y=1\) we obtain \(c=0 .\) The solution is \\[y=1+\frac{1}{10} \tan \frac{x}{10}.\\]

(a) Solving \(v_{t}=\sqrt{m g / k}\) for \(k\) we obtain \(k=m g / v_{t}^{2} .\) The differential equation then becomes \\[ m \frac{d v}{d t}=m g-\frac{m g}{v_{t}^{2}} v^{2} \\] or \(\frac{d v}{d t}=g\left(1-\frac{1}{v_{t}^{2}} v^{2}\right)\) Separating variables and integrating gives \\[ v_{t} \tanh ^{-1} \frac{v}{v_{t}}=g t+c_{1} \\] The initial condition \(v(0)=0\) implies \(c_{1}=0,\) so \\[ v(t)=v_{t} \tanh \frac{g t}{v_{t}} \\] We find the distance by integrating: \\[ s(t)=\int v_{t} \tanh \frac{g t}{v_{t}} d t=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)+c_{2} \\] The initial condition \(s(0)=0\) implies \(c_{2}=0,\) so \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right) \\] In 25 seconds she has fallen \(20,000-14,800=5,200\) feet. Using a CAS to solve \\[ 5200=\left(v_{t}^{2} / 32\right) \ln \left(\cosh \frac{32(25)}{v_{t}}\right) \\] for \(v_{t}\) gives \(v_{t} \approx 271.711 \mathrm{ft} / \mathrm{s} .\) Then \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)=2307.08 \ln (\cosh 0.117772 t) \\] (b) \(\operatorname{At} t=15, s(15)=2,542.94 \mathrm{ft}\) and \(v(15)=s^{\prime}(15)=256.287 \mathrm{ft} / \mathrm{sec}.\)
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