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Chapter 3: Problem 15

Identifying \(P(x)=2(1+x) /\left(1-2 x-x^{2}\right)\) we have $$y_{2}=(x+1) \int \frac{e^{-\int 2(1+x) d x /\left(1-2 x-x^{2}\right)}}{(x+1)^{2}} d x=(x+1) \int \frac{e^{\ln \left(1-2 x-x^{2}\right)}}{(x+1)^{2}} d x$$ $$=(x+1) \int \frac{1-2 x-x^{2}}{(x+1)^{2}} d x=(x+1) \int\left[\frac{2}{(x+1)^{2}}-1\right] d x$$ $$=(x+1)\left[-\frac{2}{x+1}-x\right]=-2-x^{2}-x$$ A second solution is \(y_{2}=x^{2}+x+2\)

Short Answer

Expert verified
The solution is verified as both forms are equivalent under certain manipulations.

Step by step solution

01

Identify the Function P(x)

We start with identifying the function given in the problem. We have \( P(x) = \frac{2(1+x)}{1-2x-x^2} \). This function will be used in calculating the exponential expression in the integral.
02

Simplify Exponential Integral Expression

The problem involves the exponential of an integral. We notice that \( \int \frac{2(1+x)}{1-2x-x^2} \, dx \) can be simplified using the property of the exponential of a logarithm. Specifically, \( e^{\ln(1 - 2x - x^2)} = 1 - 2x - x^2 \). This simplification allows us to directly substitute into the integral.
03

Simplify the Integral

We rewrite the integral as \( \int \frac{1-2x-x^2}{(x+1)^2} \, dx \). By performing polynomial division or recognizing the structure, it simplifies to \( \int \left( \frac{2}{(x+1)^2} - 1 \right) \, dx \). This is a more familiar form that we can integrate term by term.
04

Integrate the Expression

We separate and integrate the expression: \( \int \frac{2}{(x+1)^2} \, dx = -\frac{2}{x+1} \) and \( \int -1 \, dx = -x \). Combine these results to obtain the integrated expression \( -\frac{2}{x+1} - x \).
05

Multiply by (x+1)

Multiply the integrated result by \((x+1)\) as per the original construction of \( y_2 \): \( (x+1)\left(-\frac{2}{x+1} - x \right) \). This simplifies to \( -2 - x - x\cdot x = -2 - x^2 - x \).
06

Identify Second Solution as Alternative Form

The given "alternative" solution is \( y_2 = x^2 + x + 2 \). By comparing \(-2 - x^2 - x\) to this form, we recognize they are equivalent up to an additive constant. The equivalence shows the relationship between the forms is consistent under certain algebraic manipulations or adjustments to constants within a differential solution context.

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Key Concepts

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

Integral Simplification
Integral simplification is a crucial aspect when dealing with differential equations, primarily because it helps in making complex integrals more manageable. In some problems, like the one given, the presence of an exponential within an integral can seem daunting initially but is often made simpler by recognizing certain mathematical properties.One common property used in simplification is the exponential of a logarithm. The integral in our problem involved the expression \( e^{\ln(1 - 2x - x^2)} \). Applying the property \( e^{\ln(a)} = a \), the expression simplifies to \( 1 - 2x - x^2 \). These types of manipulations are critical, as they transform otherwise convoluted problems into more straightforward algebraic forms.
Recognizing patterns such as these can significantly reduce the work needed to solve integrals in many differential equations.
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent and are written in the form \( b^x \). They are widely encountered in calculus and differential equations due to their distinct properties. The problem tackled here showcases the integration of exponentials, specifically the transformation of an expression \( e^{\int f(x) dx} \) using the natural logarithm property.
When the expression within an exponential is a logarithm, it can sometimes be simplified to a polynomial. Recognizing and applying properties of exponential and logarithmic functions is vital for simplifying differential equations. These functions are governing factors in the growth, decay, and distribution phenomena, making them prevalent in various branches of science and engineering.
Polynomial Division
Polynomial division, akin to long division in arithmetic, is a method used to simplify complex polynomial expressions. It is a key technique in solving integrals where the integrand can be decomposed into simpler components.In our problem, the integral \( \int \frac{1-2x-x^2}{(x+1)^2} \, dx \) was simplified through polynomial division. Recognizing that \( \frac{1-2x-x^2}{(x+1)^2} \) could be rewritten as \( \frac{2}{(x+1)^2} - 1 \) immensely eases the integration process. Breaking down an integrand into manageable bits alters it from a single complex entity into simpler terms that can be tackled individually, often by direct integration.
Understanding and applying polynomial division can be greatly beneficial in calculus, in particular when dealing with rational functions where numerators and denominators are polynomials.

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Most popular questions from this chapter

The solution of the initial-value problem \\[ x^{\prime \prime}+\omega^{2} x=0, \quad x(0)=0, x^{\prime}(0)=v_{0}, \omega^{2}=10 / m \\] is \(x(t)=\left(v_{0} / \omega\right) \sin \omega t .\) To satisfy the additional boundary condition \(x(1)=0\) we require that \(\omega=n \pi\) \(n=1,2,3, \ldots,\) The eigenvalues \(\lambda=\omega^{2}=n^{2} \pi^{2}\) and eigenfunctions of the problem are then \(x(t)=\) \(\left(v_{0} / n \pi\right) \sin n \pi t . \quad\) Using \(\omega^{2}=10 / m\) we find that the only masses that can pass through the equilibrium position at \(t=1\) are \(m_{n}=10 / n^{2} \pi^{2} .\) Note for \(n=1,\) the heaviest mass \(m_{1}=10 / \pi^{2}\) will not pass through the equilibrium position on the interval \(0 < t<1\) (the period of \(x(t)=\left(v_{0} / \pi\right) \sin \pi t\) is \(T=2,\) so on \(0 \leq t \leq 1\) its graph passes through \(x=0\) only at \(t=0\) and \(t=1\) ). Whereas for \(n>1\), masses of lighter weight will pass through the equilibrium position \(n-1\) times prior to passing through at \(t=1 .\) For example, if \(n=2,\) the period of \(x(t)=\left(v_{0} / 2 \pi\right) \sin 2 \pi t\) is \(2 \pi / 2 \pi=1,\) the mass will pass through \(x=0\) only once \(\left(t=\frac{1}{2}\right)\) prior to \(t=1 ;\) if \(n=3,\) the period of \(x(t)=\left(v_{0} / 3 \pi\right) \sin 3 \pi t\) is \(\frac{2}{3},\) the mass will pass through \(x=0\) twice \(\left(t=\frac{1}{3} \text { and } t=\frac{2}{3}\right)\) prior to \(t=1 ;\) and so on.

Solving \(\frac{1}{4} q^{\prime \prime}+20 q^{\prime}+300 q=0\) we obtain \(q(t)=c_{1} e^{-20 t}+c_{2} e^{-60 t} .\) The initial conditions \(q(0)=4\) and \(q^{\prime}(0)=0\) imply \(c_{1}=6\) and \(c_{2}=-2 .\) Thus \\[q(t)=6 e^{-20 t}-2 e^{-60 t}\\] Setting \(q=0\) we find \(e^{40 t}=1 / 3\) which implies \(t<0 .\) Therefore the charge is not 0 for \(t \geq 0\).

Using the general solution \(y=c_{1} \cos \sqrt{3} x+c_{2} \sin \sqrt{3} x+2 x,\) the boundary conditions \(y(0)+y^{\prime}(0)=0, y(1)=0\) yield the system $$\begin{array}{r} c_{1}+\sqrt{3} c_{2}+2=0 \\ c_{1} \cos \sqrt{3}+c_{2} \sin \sqrt{3}+2=0 \end{array}$$ Solving gives $$c_{1}=\frac{2(-\sqrt{3}+\sin \sqrt{3})}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}} \quad \text { and } \quad c_{2}=\frac{2(1-\cos \sqrt{3})}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}$$ Thus, $$y=\frac{2(-\sqrt{3}+\sin \sqrt{3}) \cos \sqrt{3} x}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}+\frac{2(1-\cos \sqrt{3}) \sin \sqrt{3} x}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}+2 x$$

Solving \(\frac{1}{2} q^{\prime \prime}+20 q^{\prime}+1000 q=0\) we obtain \(q_{c}(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The steady- state charge has the form \(q_{p}(t)=A \sin 60 t+B \cos 60 t+C \sin 40 t+D \cos 40 t .\) Substituting into the differential equation we find $$\begin{aligned}(-1600 A-2400 B) & \sin 60 t+(2400 A-1600 B) \cos 60 t \\ +&(400 C-1600 D) \sin 40 t+(1600 C+400 D) \cos 40 t\end{aligned}$$ $$=200 \sin 60 t+400 \cos 40 t$$ Equating coefficients we obtain \(A=-1 / 26, B=-3 / 52, C=4 / 17,\) and \(D=1 / 17 .\) The steady-state charge is \\[q_{p}(t)=-\frac{1}{26} \sin 60 t-\frac{3}{52} \cos 60 t+\frac{4}{17} \sin 40 t+\frac{1}{17} \cos 40 t\\] and the steady-state current is \\[i_{p}(t)=-\frac{30}{13} \cos 60 t+\frac{45}{13} \sin 60 t+\frac{160}{17} \cos 40 t-\frac{40}{17} \sin 40 t\\].

We have \(y_{c}=c_{1} \cos x+c_{2} \sin x\) and we assume \(y_{p}=A^{2}+B x+C .\) Substituting into the differential equation we find \(A=1, B=0,\) and \(C=-1 .\) Thus \(y=c_{1} \cos x+c_{2} \sin x+x^{2}-1 .\) From \(y(0)=5\) and \(y(1)=0\) we obtain $$\begin{aligned} &c_{1}-1=5\\\ &(\cos 1) c_{1}+(\sin 1) c_{2}=0 \end{aligned}$$ Solving this system we find \(c_{1}=6\) and \(c_{2}=-6\) cot \(1 .\) The solution of the boundary-value problem is $$y=6 \cos x-6(\cot 1) \sin x+x^{2}-1$$
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