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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