12.5: Problems
- Page ID
- 91038
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Find all of the solutions of the first order differential equations. When an initial condition is given, find the particular solution satisfying that condition.
- \(\frac{d y}{d x}=\frac{e^{x}}{2 y}\).
- \(\frac{d y}{d t}=y^{2}\left(1+t^{2}\right), y(0)=1\).
- \(\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{x}\).
- \(x y^{\prime}=y(1-2 y), \quad y(1)=2\).
- \(y^{\prime}-(\sin x) y=\sin x\).
- \(x y^{\prime}-2 y=x^{2}, y(1)=1\).
- \(\frac{d s}{d t}+2 s=s t^{2}, \quad, s(0)=1 .\)
- \(x^{\prime}-2 x=t e^{2 t}\).
- \(\frac{d y}{d x}+y=\sin x, y(0)=0\).
- \(\frac{d y}{d x}-\frac{3}{x} y=x^{3}, y(1)=4\).
Consider the differential equation \[\frac{d y}{d x}=\frac{x}{y}-\frac{x}{1+y} .\nonumber\]
- Find the 1-parameter family of solutions (general solution) of this equation.
- Find the solution of this equation satisfying the initial condition \(y(0)=1\). Is this a member of the 1-parameter family?
Identify the type of differential equation. Find the general solution and plot several particular solutions. Also, find the singular solution if one exists.
- \(y=x y^{\prime}+\frac{1}{y^{\prime}}\).
- \(y=2 x y^{\prime}+\ln y^{\prime}\).
- \(y^{\prime}+2 x y=2 x y^{2}\).
- \(y^{\prime}+2 x y=y^{2} e^{x^{2}}\).
Find all of the solutions of the second order differential equations. When an initial condition is given, find the particular solution satisfying that condition.
- \(y^{\prime \prime}-9 y^{\prime}+20 y=0\).
- \(y^{\prime \prime}-3 y^{\prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1\).
- \(8 y^{\prime \prime}+4 y^{\prime}+y=0, \quad y(0)=1, \quad y^{\prime}(0)=0\).
- \(x^{\prime \prime}-x^{\prime}-6 x=0\) for \(x=x(t)\).
Verify that the given function is a solution and use Reduction of Order to find a second linearly independent solution.
- \(x^{2} y^{\prime \prime}-2 x y^{\prime}-4 y=0, \quad y_{1}(x)=x^{4}\).
- \(x y^{\prime \prime}-y^{\prime}+4 x^{3} y=0, \quad y_{1}(x)=\sin \left(x^{2}\right)\).
Prove that \(y_{1}(x)=\sinh x\) and \(y_{2}(x)=3 \sinh x-2 \cosh x\) are linearly independent solutions of \(y^{\prime \prime}-y=0\). Write \(y_{3}(x)=\cosh x\) as a linear combination of \(y_{1}\) and \(y_{2}\).
Consider the nonhomogeneous differential equation \(x^{\prime \prime}-3 x^{\prime}+2 x=6 e^{3 t}\).
- Find the general solution of the homogenous equation.
- Find a particular solution using the Method of Undetermined Coefficients by guessing \(x_{p}(t)=A e^{3 t}\).
- Use your answers in the previous parts to write down the general solution for this problem.
Find the general solution of the given equation by the method given.
- \(y^{\prime \prime}-3 y^{\prime}+2 y=10\). Method of Undetermined Coefficients.
- \(y^{\prime \prime}+y^{\prime}=3 x^{2}\). Variation of Parameters.
Use the Method of Variation of Parameters to determine the general solution for the following problems.
- \(y^{\prime \prime}+y=\tan x\).
- \(y^{\prime \prime}-4 y^{\prime}+4 y=6 x e^{2 x}\).
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.
Find all of the solutions of the second order differential equations for \(x>0\).. When an initial condition is given, find the particular solution satisfying that condition.
- \(x^{2} y^{\prime \prime}+3 x y^{\prime}+2 y=0\).
- \(x^{2} y^{\prime \prime}-3 x y^{\prime}+3 y=0\).
- \(x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0\).
- \(x^{2} y^{\prime \prime}-2 x y^{\prime}+3 y=0\).
- \(x^{2} y^{\prime \prime}+3 x y^{\prime}-3 y=x^{2}\).