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Chapter 2: Problem 2

Find all the solutions of the second-order differential equations. When an initial condition is given, find the particular solution satisfying that condition. a. \(y^{\prime \prime}-9 y^{\prime}+20 y=0\). b. \(y^{\prime \prime}-3 y^{\prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1\). c. \(x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0, \quad x>0\). d. \(x^{2} y^{\prime \prime}-2 x y^{\prime}+3 y=0, \quad x>0\)

Short Answer

Expert verified
a) \(y = c_1 e^{4x} + c_2 e^{5x}\). b) \(y = e^{1.5x}( A\cos(\sqrt{7/4}x) + B\sin(\sqrt{7/4}x))\), with A and B are determined by initial conditions. c) \(y = c_1 x^{m_1} + c_2 x^{m_2}\), where \(m_1\) and \(m_2\) are roots of the characteristic equation. d) Similar to c, with different roots.

Step by step solution

01

Solving equation a

To find the solution of the second-order differential equation \(y^{\prime \prime}-9 y^{\prime}+20 y=0\), we start by solving the characteristic equation \(m^2 - 9m + 20 = 0\). Solving this quadratic equation gives the roots \(m=4\) and \(m=5\). Therefore, the general solution of the second-order differential equation is \(y=c_1 e^{4x} + c_2 e^{5x}\) where \(c_1\) and \(c_2\) are arbitrary constants.
02

Solving equation b

Start by solving the given initial value problem's characteristic equation, \(y^{\prime \prime}-3 y^{\prime}+4 y=0 \). The characteristic equation is \(m^2 - 3m + 4 = 0\), which has roots \(m=1.5 \pm \sqrt{7/4}\). The two solutions are complex and conjugate, so the general solution is in the form \((y= e^{mt}( A\cos(nt) + B\sin(nt)))\). Here, \(m\) is the real part and \(n\) is the imaginary part of the roots. To solve for particular solution that satisfies \(y(0)=0\) and \(y^{\prime}(0)=1\), plug those conditions into your general solution and its derivative.
03

Solving equations c and d

These are examples of the Cauchy-Euler type of differential equations. Equation c, \(x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0\) has the general solution \(y = c_1 x^{m_1} + c_2 x^{m_2}\) where \(m_1\) and \(m_2\) are roots of the characteristic equation \(m(m-1) + 5m + 4 = 0\). Similarly, equation d, \(x^{2} y^{\prime \prime}-2 x y^{\prime}+3 y=0\), has the same form, with roots of \(m(m-1) - 2m + 3 = 0\).

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Most popular questions from this chapter

Consider the nonhomogeneous differential equation \(x^{\prime \prime}-3 x^{\prime}+2 x=6 e^{3 t}\). a. Find the general solution of the homogenous equation. b. Find a particular solution using the Method of Undetermined Coefficients by guessing \(x_{p}(t)=A e^{3 t}\). c. Use your answers in the previous parts to write the general solution for this problem.

Find all the solutions of the first-order differential equations. When an initial condition is given, find the particular solution satisfying that condition. a. \(\frac{d y}{d x}=\frac{e^{x}}{2 y}\). b. \(\frac{d y}{d t}=y^{2}\left(1+t^{2}\right), y(0)=1\). c. \(\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{x}\) d. \(x y^{\prime}=y(1-2 y), \quad y(1)=2\) e. \(y^{\prime}-(\sin x) y=\sin x\) f. \(x y^{\prime}-2 y=x^{2}, y(1)=1\) g. \(\frac{d s}{d t}+2 s=s t^{2}, \quad s(0)=1\) h. \(x^{\prime}-2 x=t e^{2 t}\) i. \(\frac{d y}{d x}+y=\sin x, y(0)=0\). j. \(\frac{d y}{d x}-\frac{3}{x} y=x^{3}, y(1)=4\)

Use the transformations relating polar and Cartesian coordinates to prove that $$ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left[x \frac{d y}{d t}-y \frac{d x}{d t}\right] $$

Instead of assuming that \(c_{1}^{\prime} y_{1}+c_{2}^{\prime} y_{2}=0\) in the derivation of the solution using Variation of Parameters, assume that \(c_{1}^{\prime} y_{1}+c_{2}^{\prime} y_{2}=h(x)\) for an arbitrary function \(h(x)\) and show that one gets the same particular solution.

Use i) Euler's Method and ii) the Midpoint Method to determine the given value of \(y\) for the following problems: a. \(\frac{d y}{d x}=2 y, y(0)=2\). Find \(y(1)\) with \(h=0.1\). b. \(\frac{d y}{d x}=x-y, y(0)=1\). Find \(y(2)\) with \(h=0.2\). c. \(\frac{d y}{d x}=x \sqrt{1-y^{2}}, y(1)=0 .\) Find \(y(2)\) with \(h=0.2\).
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