Problem 37

Question

Write the differential equation as $$\frac{d v}{d x}+\frac{1}{x} v=32 v^{-1},$$ and let $u=v^{2}$ or $v=u^{1 / 2} .$ Then $$\frac{d v}{d x}=\frac{1}{2} u^{-1 / 2} \frac{d u}{d x},$$ and substituting into the differential equation, we have $$\frac{1}{2} u^{-1 / 2} \frac{d u}{d x}+\frac{1}{x} u^{1 / 2}=32 u^{-1 / 2} \quad \text { or } \quad \frac{d u}{d x}+\frac{2}{x} u=64.$$ The latter differential equation is linear with integrating factor $e^{\int(2 / x) d x}=x^{2},$ so $$\frac{d}{d x}\left[x^{2} u\right]=64 x^{2}$$ and $$x^{2} u=\frac{64}{3} x^{3}+c \quad \text { or } \quad v^{2}=\frac{64}{3} x+\frac{c}{x^{2}}.$$

Step-by-Step Solution

Verified

Answer

The solution is $ v^2 = \frac{64}{3} x + \frac{C}{x^2} $.

1Step 1: Substitute Expression for v

Given that $ u = v^2 $, substitute $ v = u^{1/2} $ into the differential equation. Thus the derivative $ \frac{d v}{d x} = \frac{1}{2} u^{-1/2} \frac{d u}{d x} $ can be substituted into the original equation.

2Step 2: Substitute into Original Equation

Replace $ \frac{d v}{d x} $ in the differential equation $ \frac{d v}{d x} + \frac{1}{x} v = 32 v^{-1} $ with $ \frac{1}{2} u^{-1/2} \frac{d u}{d x} $, resulting in the equation: $ \frac{1}{2} u^{-1/2} \frac{d u}{d x} + \frac{1}{x} u^{1/2} = 32 u^{-1/2} $.

3Step 3: Simplify the Equation

Multiply through by $ 2u^{1/2} $ to eliminate the fractional exponent, leading to the simplified form $ \frac{d u}{d x} + \frac{2}{x} u = 64 $.

4Step 4: Find Integrating Factor

Identify the linear differential equation $ \frac{d u}{d x} + \frac{2}{x} u = 64 $. The integrating factor is found by evaluating $ e^{\int \frac{2}{x} \, dx} = x^2 $.

5Step 5: Multiply by Integrating Factor

Multiply the entire differential equation by the integrating factor $ x^2 $ to transform it into $ \frac{d}{d x} (x^2 u) = 64 x^2 $.

6Step 6: Integrate Both Sides

Integrate both sides with respect to $ x $. The left-hand side integrates to $ x^2 u $ and the right-hand side integrates to $ \frac{64}{3} x^3 + C $.

7Step 7: Solve for u

Equating the two sides gives $ x^2 u = \frac{64}{3} x^3 + C $. Solving for $ u $ gives $ u = \frac{64}{3} x + \frac{C}{x^2} $.

8Step 8: Rewrite in terms of v

Since $ u = v^2 $, substitute back to express in terms of $ v $: $ v^2 = \frac{64}{3} x + \frac{C}{x^2} $.

Key Concepts

The Integrating FactorUnderstanding the Substitution MethodDifferential Equation Simplification

The Integrating Factor

In solving linear differential equations, the integrating factor is a powerful tool. It is used to transform the equation into an easily integrable form.
The process involves finding a function that, when multiplied to the whole equation, converts it into an exact derivative. Consider a linear differential equation of the form $ \frac{du}{dx} + P(x)u = Q(x) $.

The integrating factor is no mystery; it is calculated using the formula $ e^{\int P(x) \, dx} $. In our specific case, where $ P(x) = \frac{2}{x} $, the integrating factor is $ x^2 $.

Multiply the entire differential equation by the integrating factor.
The left side becomes: $ \frac{d}{dx}(x^2 u) $.
This transformation makes it easy to integrate both sides, further simplifying the solution.

Understanding and using the integrating factor correctly helps in solving differential equations effectively.

Understanding the Substitution Method

The substitution method is a technique used in differential equations to simplify complicated relationships. It involves replacing a variable with another expression to ease the manipulation of the equation.
For instance, in our original problem, $ v $ is replaced with $ u^{1/2} $, where $ u = v^2 $.

This substitution helps in simplifying the derivatives and equations. Here's how it works:

First, acknowledge the relationship between variables. For example, if $ v = u^{1/2} $, then $ u = v^2 $.
Replace $ v $ and its derivatives in the equation with $ u $ and its derivatives. This step makes the equation more manageable.
This method often turns a non-linear equation into a linear one, which is much simpler to solve.

Employing substitution can make complex equations more approachable and easier to solve.

Differential Equation Simplification

Simplifying a differential equation is a crucial technique for finding solutions efficiently. This involves manipulating the equation to a form that's easier to integrate or differentiate.
Initially, our equation was non-linear and more complex, hence the need for simplification.

Through techniques like substitution, mentioned earlier, we simplified $ \frac{dv}{dx} + \frac{1}{x}v = 32v^{-1} $ into a more manageable equation: $ \frac{du}{dx} + \frac{2}{x}u = 64 $.

Begin by identifying terms that can be substituted or simplified.
Apply mathematical transformations to rewrite the differential equation into a linear form.
Linear equations are generally easier to handle as they can be solved systematically using integrating factors or direct integration.

Mastering these simplification methods allows quicker and more efficient solutions to challenging differential equations.

Problem 36

Problem 37

Other exercises in this chapter

Problem 36

For $y^{\prime}+e^{x} y=1$ an integrating factor is $e^{e^{x}}$. Thus \\[ \frac{d}{d x}\left[e^{e^{x}} y\right]=e^{e^{x}} \quad \text { and } \quad e^{e^{x}

View solution

Problem 36

We note that $\left(N_{x}-M_{y}\right) / M=-3 / y,$ so an integrating factor is $e^{-3} \int d y / y=1 / y^{3} .$ Let \(M=\left(y^{2}+x y^{3}\right) / y^{3}

View solution

Problem 37

Separating variables, we have \\[\frac{d y}{y-y^{3}}=\frac{d y}{y(1-y)(1+y)}=\left(\frac{1}{y}+\frac{1 / 2}{1-y}-\frac{1 / 2}{1+y}\right) d y=d x\\]. Integratin

View solution

Problem 37

An integrating factor for $y^{\prime}-2 x y=1$ is $e^{-x^{2}} .$ Thus \\[ \begin{aligned} \frac{d}{d x}\left[e^{-x^{2}} y\right] &=e^{-x^{2}} \\ e^{-x^{2}}

View solution