/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Solve the following differential... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 11

Solve the following differential equations: (i) \(y^{\prime}-y \ln 2=2^{\sin x}(\cos x-1) \ln 2, y\) being bounded when \(\mathrm{x} \rightarrow \infty\). (ii) \(y^{\prime} \sin x-y \cos x=-\frac{\sin ^{2} x}{x^{2}}, y \rightarrow 0\) as \(x \rightarrow \infty\) (iii) \(x^{2} y^{\prime} \cos \frac{1}{x}-y \sin \frac{1}{x}=-1, y \rightarrow 1\) as \(x \rightarrow \infty\). (iv) \(x^{2} y^{\prime}+y=\left(x^{2}+1\right) e^{x}, y \rightarrow 1\) as \(x \rightarrow \infty\)

Short Answer

Expert verified
Based on the given step by step solution, create a short answer question: Question: Solve the following first order, linear, inhomogeneous differential equation: \(x^2 y^{\prime}+y = \left(x^2+1\right) e^{x}\). Answer: To solve this differential equation, we find the integrating factor: \(IF = e^{\int \frac{1}{x^2} \,dx} = x\). Multiply the equation by the integrating factor: \(x^3 y^{\prime}+x^2 y = x^2(x^2+1)e^x\). The left side is the derivative of \(y \cdot x^2\), so we have \((yx^2)^{\prime} = x^2(x^2+1)e^x\). Integrating both sides: \(yx^2 = \int x^2(x^2+1)e^x \, dx + C\). Now divide by \(x^2\) to find the solution for y: \(y(x) = \frac{1}{x^2} \left[\int x^2(x^2+1)e^x \, dx + C \right]\). The integral doesn't have a closed form solution, and hence we can't apply the boundary condition.

Step by step solution

01

Problem (i) - Differential equation

Given the equation \(y^{\prime}-y \ln 2=2^{\sin x}(\cos x-1) \ln 2\). This equation is a first order, linear, inhomogeneous differential equation. An equation of the form: \(y^{\prime} + p(x)y = q(x)\)
02

Problem (i) - Solving the differential equation

The integrating factor is \(IF = e^{\int -\ln 2 \,dx} = e^{-x \ln 2}\). Now multiply the given equation by the integrating factor: \(e^{-x \ln 2}(y^{\prime}-y \ln 2)=e^{-x \ln 2}2^{\sin x}(\cos x-1) \ln 2\) The left side is the derivative of \(y \cdot e^{-x \ln 2}\): \((y e^{-x \ln 2})^{\prime} = e^{-x \ln 2}2^{\sin x}(\cos x-1) \ln 2\) Now, integrate both sides: \(y e^{-x \ln 2} = \int e^{-x \ln 2}2^{\sin x}(\cos x-1) \ln 2 \, dx + C\) To find the solution for y, multiply by the inverse of the integrating factor: \( e^{x \ln 2}\) \(y(x) = e^{x \ln 2} \int e^{-x \ln 2}2^{\sin x}(\cos x-1) \ln 2 \, dx + Ce^{x \ln 2}\) The bounded condition is not used in this case as the integral cannot be evaluated explicitly.
03

Problem (ii) - Differential equation

Given the equation \(y^{\prime} \sin x-y \cos x=-\frac{\sin^2 x}{x^2} \,,\) with \( y \rightarrow 0\) as \(x \rightarrow \infty\). This equation is also a first order, linear, inhomogeneous differential equation.
04

Problem (ii) - Solving the differential equation

The integrating factor is \(IF = e^{\int -\cot x \,dx} = e^{-\ln \sin x} = \csc x\). Multiply the given equation by the integrating factor: \(\csc x (y^{\prime} \sin x - y \cos x) = -\frac{\sin^2 x}{x^2} \csc x\) The left side is the derivative of \(y \cdot \csc x\): \((y \csc x)^{\prime} = -\frac{\sin^2 x}{x^2} \csc x\) Now integrate both sides: \(y \csc x = \int -\frac{\sin^2 x}{x^2} \csc x \, dx + C\) Multiply by \(\sin x \) to find the solution for y: \(y(x) = \int -\frac{\sin^3 x}{x^2} \, dx + C \sin x\) The boundary condition can't be used since the integral can not be evaluated explicitly.
05

Problem (iii) - Differential equation

Given the equation \(x^2 y^{\prime} \cos \frac{1}{x}-y \sin \frac{1}{x} = -1 \,,\) with \( y \rightarrow 1\) as \(x \rightarrow \infty\). The equation is similar to the previous two problems which means it is a first order, linear, inhomogeneous differential equation.
06

Problem (iii) - Solving the differential equation

The integrating factor is \(IF = e^{\int -\frac{1}{x}\cot \frac{1}{x} \,dx} = x\). Multiply the given equation by the integrating factor: \(x^3 y^{\prime} \cos \frac{1}{x} - x^2 y \sin \frac{1}{x} = -x\) The left side is the derivative of \(y \cdot x^2 \sin \frac{1}{x}\): \((y \cdot x^2 \sin \frac{1}{x})^{\prime} = -x\) Now integrate both sides: \(y \cdot x^2 \sin \frac{1}{x} = \int -x \, dx + C = -\frac{1}{2}x^2 + C\) Now divide by \(x^2 \sin \frac{1}{x}\) to find the solution for y: \(y(x) = -\frac{1}{2} + \frac{C}{x^2 \sin \frac{1}{x}}\) Applying the boundary condition \(y \rightarrow 1\) as \(x \rightarrow \infty\), we get \(C= \frac{1}{2}\). So, the solution is: \(y(x) = -\frac{1}{2} + \frac{1}{2x^2 \sin \frac{1}{x}}\)
07

Problem (iv) Differential equation

Given the equation \(x^2 y^{\prime}+y = \left(x^2+1\right) e^{x}\), with \(y \rightarrow 1\) as \(x \rightarrow \infty\). Like the previous problems, this is also a first order, linear, inhomogeneous differential equation.
08

Problem (iv) Solving the differential equation

The integrating factor is \(IF = e^{\int \frac{1}{x^2} \,dx} = x\). Multiply the given equation by the integrating factor: \(x^3 y^{\prime}+x^2 y = x^2(x^2+1)e^x\) The left side is the derivative of \(y \cdot x^2\): \((yx^2)^{\prime} = x^2(x^2+1)e^x\) Integrate both sides: \(yx^2 = \int x^2(x^2+1)e^x \, dx + C\) Now divide by \(x^2\) to find the solution for y: \(y(x) = \frac{1}{x^2} \left[\int x^2(x^2+1)e^x \, dx + C \right]\) This integral doesn't have a closed form solution, and hence we can't apply the boundary condition.

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Most popular questions from this chapter

Find all solutions of \(y^{\prime} \sin x+y \cos x=1\) on the interval \((0, \pi)\). Prove that exactly one of these solutions has a finite limit as \(\mathrm{x} \rightarrow 0\), and another has a finite limit as \(\mathrm{x} \rightarrow \pi\).

Prove that the differential equation of the confocal conics \(\frac{\mathrm{x}^{3}}{\mathrm{a}^{2}+\lambda}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda}=1\), is \(\mathrm{xyp}^{2}+\left(\mathrm{x}^{2}-\mathrm{y}^{2}-\right.\) \(\left.a^{2}+b^{2}\right) p-x y=0\) Show that this coincides with the differential equation of the orthogonal curves, and interpret the result.

Show that the half-life \(\mathrm{H}\) of a radioactive substance can be obtained from two measurements \(\mathrm{y}_{1}=\mathrm{y}\left(\mathrm{t}_{1}\right)\) and \(\mathrm{y}_{2}=\mathrm{y}\left(\mathrm{t}_{2}\right)\) of the amount present at times \(t_{1}\) and \(t_{2}\) by the formula \(H=\left[\left(t_{2}-t_{1}\right) \ln 2\right] / \ln \left(y_{1} / y_{2}\right)\)

A function of \(\mathrm{x}\) is a solution of a differential equation if it and its derivatives make the equation true. For what value (or values) of \(\mathrm{m}\) is \(\mathrm{y}=\mathrm{e}^{\mathrm{mx}}\) a solution of \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-3 \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0\) ?

Find the solution of the equation \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}|\ln \mathrm{y}|^{\alpha}\), \(\alpha>0\) satisfying the initial condition \(\mathrm{y}(0)=0\). For what values of \(\alpha\) has the problem a unique solution?
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