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Chapter 31: Problem 36

Find the indicated particular solutions of the given differential equations. $$\begin{aligned} &f(x) d y+2 y f^{\prime}(x) d x=f(x) f^{\prime}(x) d x ; \quad f(x)=-1 \text { when }\\\ &y=3 \end{aligned}$$

Short Answer

Expert verified
The particular solution is \(y = \frac{f(x)}{2} - \frac{7}{2f(x)^2}\).

Step by step solution

01

Simplify the given differential equation

Given the differential equation, \(f(x) dy + 2y f'(x) dx = f(x) f'(x) dx\), let's rearrange the terms. Move like terms to the same side:\[f(x) dy = f(x) f'(x) dx - 2y f'(x) dx\] \(f(x) dy = f'(x) (f(x) - 2y) dx\). This gives a simplified differential form.
02

Separate variables for integration

The equation \(f(x) dy = f'(x) (f(x) - 2y) dx\) can be separated as follows: \[\frac{dy}{f(x) - 2y} = \frac{f'(x)}{f(x)} dx\] Now we've separated the variables for integration on both sides.
03

Integrate both sides

Integrating both sides of the equation:- The left side, \(\int \frac{dy}{f(x) - 2y}\) is a standard integral of type \(\int \frac{1}{ax+b} dx\), resulting in:\[-\frac{1}{2} \ln |f(x) - 2y| + C_1\]- The right side, \(\int \frac{f'(x)}{f(x)} dx\), results in:\[ \ln |f(x)| + C_2\]Thus, equating both gives:\[-\frac{1}{2} \ln |f(x) - 2y| = \ln |f(x)| + C\]where \(C = C_2 - C_1\).
04

Exponentiate to solve for \(y\)

Exponentiating both sides to remove the logarithm yields:\[|f(x) - 2y|^{-1/2} = K |f(x)|\]where \(K = e^C\).Squaring both sides gives:\[f(x) - 2y = \frac{K^2}{f(x)^2}\]. Rearrange to solve for \(y\):\[y = \frac{f(x)}{2} - \frac{K^2}{2f(x)^2}\].
05

Use initial condition to find particular solution

We are given that \(y = 3\) when \(f(x) = -1\). Substitute these values into the equation:\[3 = \frac{-1}{2} - \frac{K^2}{2(-1)^2}\]Solving gives: \[3 = -\frac{1}{2} - \frac{K^2}{2} \\frac{7}{2} = \frac{K^2}{2} \K^2 = 7\].Substitute back to find \(y\):\[y = \frac{f(x)}{2} - \frac{7}{2f(x)^2}\].

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Key Concepts

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

Separation of Variables
When dealing with differential equations, separating variables is a technique that makes solving them simpler. In this exercise, the equation is initially given in a form where the functions of the variables are intermixed:
  • Initially, we have: \[f(x) \frac{dy}{dx} = f'(x) (f(x) - 2y)\]
  • To separate the variables, rearrange terms so the equation is expressed as the product of two distinct functions, one in terms of \(x\) and the other in terms of \(y\):\[\frac{dy}{f(x) - 2y} = \frac{f'(x)}{f(x)} dx\]
Each side of the equation now only involves one variable, enabling us to integrate each part independently. This separation simplifies further steps by isolating variables, making the equation straightforward to solve. Separating variables is often the first step to solve such equations since it prepares the equation for integration.
Integration Techniques
Integration is the process of finding a function from its derivative, and it's a crucial step here. Once we've separated the variables, we integrate each side of the equation:
  • The left side involves an integral in terms of \(y\):\[\int \frac{dy}{f(x) - 2y} = -\frac{1}{2} \ln |f(x) - 2y| + C_1\]This integral uses the substitution method akin to the standard integration formula \(\int \frac{1}{ax+b} dx\).
  • For the right side, we have an integral in terms of \(x\):\[\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C_2\]This is solved using another standard integration rule, leading us to the solution involving logarithmic functions.
Both integrations introduce constants \(C_1\) and \(C_2\), which we can combine into a single constant \(C\) later on. The resulting equation in terms of both \(x\) and \(y\) then allows us to exponentiate both sides to solve for \(y\). This step is essential for further transformation of the solution into a usable form.
Initial Condition
The initial condition is crucial in finding a particular solution to a differential equation. Here, we know that when \(f(x) = -1\), \(y = 3\). This information helps us find the exact value of any constants introduced during integration.
  • We substitute the initial condition into the equation found after integration and exponentiation:\[3 = \frac{-1}{2} - \frac{K^2}{2(-1)^2}\]
  • Solving this equation gives\(K^2 = 7\), thus pinning down the constant in the equation:\[y = \frac{f(x)}{2} - \frac{7}{2f(x)^2}\]
By applying the initial condition, the integration constant \(C\) ultimately affects the specific form of our particular solution. This particular solution not only fits the general solution but also satisfies the specific numerical condition provided, offering a specific outcome rooted in the broader family of solutions for the differential equation. Initial conditions personalize the equation, tailoring it for specific scenarios or datasets.

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Most popular questions from this chapter

Solve the given differential equations. $$4 y^{\prime \prime}-4 y^{\prime}+y=4 e^{x / 2}$$

Solve the given problems. The displacement \(y\) (in \(\mathrm{cm}\) ) of an object at the end of a spring is described by the equation \(d^{2} y / d t^{2}+4 d y / d t+4 y=0,\) where \(t\) is the time (in \(\mathrm{s}\) ) if \(f(0)=0\) and \(f(1)=0.50 \mathrm{cm} .\) Solve for \(y=f(t)\).

Find the particular solutions of the given differential equations that satisfy the given conditions. \(2 D^{2} y+5 D y=0 ; \quad y=0\) when \(x=0,\) and \(y=2\) when \(x=1\)

Solve the given problems by solving the appropriate differential equation. Assume that the rate at which highway construction increases is directly proportional to the total mileage \(M\) of all highways already completed at time \(t\) (in years). Solve for \(M\) as a function of \(t\) if \(M=5250 \mathrm{mi}\) for a certain county when \(t=0\) and \(M=5460 \mathrm{mi}\) for \(t=2.00\) years.

Solve the given differential equations. $$D^{2} y+2 D y=8 x+e^{-2 x}$$
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