Solution. The corresponding characteristic equations are

\[ \dot {x}=x,\qquad \dot {y}=1,\qquad \dot {z}=0. \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=u_0(r). \]

The general solutions of the characteristic equations are

\[ x(t)=c_1\e ^t,\qquad y(t)=t+c_2,\qquad z(t)=c_3, \]

for constants (in \(t\)) \(c_1\), \(c_2\) and \(c_3\). Determining the constants from the initial conditions gives

\[ x(r,t)=r\e ^t,\qquad y(r,t)=t,\qquad z(r,t)=u_0(r). \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(t=y\), \(r=x\e ^{-y}\). Substituting in \(z\) we obtain

\[ u(x,y)=u_0(x\e ^{-y}). \]

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Solution. The corresponding characteristic equations are

\[ \dot {x}=1,\qquad \dot {y}=x^2,\qquad \dot {z}=0, \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=u_0(r). \]

The solution of this initial value problem is

\[ x(r,t)=t+r,\qquad y(r,t)=\frac {1}{3}(t+r)^3-\frac {1}{3}r^3,\qquad z(r,t)=u_0(r). \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(r=(x^3-3y)^{1/3}\) and \(t=x-(x^3-3y)^{1/3}\). Substituting in \(z\) we obtain

\[ u(x,y)=u_0(\sqrt [3]{x^3-3y}). \]

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Solution. The corresponding characteristic equations are

\[ \dot {x}=y,\qquad \dot {y}=x,\qquad \dot {z}=0, \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=u_0(r). \]

The solution of this initial value problem is

\[ x(r,t)=r\cosh (t),\qquad y(r,t)=r\sinh (t),\qquad z(r,t)=u_0(r). \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(r=\pm \sqrt {x^2-y^2}\) and \(t=\arctanh (y/x)\). Substituting in \(z\) we obtain (taking into account the initial condition)

\[ u(x,y)= \begin {cases} u_0(\sqrt {x^2-y^2})&x>0\\ u_0(-\sqrt {x^2-y^2})&x<0. \end {cases} \]

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Solution. The corresponding characteristic equations are

\[ \dot {x}=y,\qquad \dot {y}=x,\qquad \dot {z}=xy^2, \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=u_0(r). \]

The solution of the initial value problem for \(x\) and \(y\) is

\[ x(r,t)=r\cosh (t),\qquad y(r,t)=r\sinh (t). \]

The differential equation for \(z\) then is

\[ \dot {z}=r^3\cosh (t)\sinh ^2(t), \]

which has general solution (recognizing the chain rule)

\[ z=\frac {1}{3}r^3\sinh ^3(t)+C. \]

Determining the constant from the initial condition gives

\[ z=\frac {1}{3}r^3\sinh ^3(t)+u_0(r). \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(r=\pm \sqrt {x^2-y^2}\) and \(r^3\sinh ^3(t)=y^3\). Substituting in \(z\) gives (taking into account the initial condition)

\[ u(x,y)= \begin {cases} \frac {1}{3}y^3+u_0(\sqrt {x^2-y^2})&x>0\\ \frac {1}{3}y^3+u_0(-\sqrt {x^2-y^2})&x<0. \end {cases} \]

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Solution. The corresponding characteristic equations are

\[ \dot {x}=x^2,\qquad \dot {y}=y^2,\qquad \dot {z}=(x-y)z, \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=1,\qquad z(r,0)=1. \]

The solution of the initial value problem for \(x\) and \(y\) is

\[ x(r,t)=\frac {-1}{t-\frac {1}{r}},\qquad y(r,t)=\frac {-1}{t-1}. \]

The differential equation for \(z\) then is

\[ \dot {z}=\left (\frac {-1}{t-\frac {1}{r}}+\frac {1}{t-1}\right )z, \]

which has general solution (by separation of variables)

\[ z=C\frac {t-1}{rt-1}. \]

Determining the constant from the initial condition gives

\[ z=\frac {t-1}{rt-1}. \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) and obtain \(t=1-\frac {1}{y}\) and \(r=\frac {xy}{xy-x+y}\). Substituting in \(z\) gives

\[ u(x,y)=\frac {xy-x+y}{y^2}. \]

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Solution. The corresponding characteristic equations are

\[ \dot {x}=x,\qquad \dot {y}=-y,\qquad \dot {z}=\frac {2(x^2-y^2)}{z^4}, \]

with initial conditions

\[ x(r,0)=r,\qquad y(r,0)=r,\qquad z(r,0)=0. \]

The solution of the initial value problem for \(x\) and \(y\) is

\[ x(r,t)=r\e ^t,\qquad y(r,t)=r\e ^{-t}. \]

The differential equation for \(z\) then is

\[ \dot {z}=\frac {4r^2\sinh (2t)}{z^4}, \]

which has general solution (by separation of variables)

\[ z=\sqrt [5]{10r^2\cosh (2t)+C}. \]

Determining the constant from the initial condition gives

\[ z=\sqrt [5]{10r^2\cosh (2t)-10r^2}. \]

We can solve for \(r\) and \(t\) in terms of \(x\) and \(y\) (noting that \(x\) and \(y\) must have the same sign) and obtain \(t=\frac {1}{2}\ln \left (\frac {x}{y}\right )\) and \(r=\pm \sqrt {xy}\). Substituting in \(z\) gives

\[ u(x,y)=\left (5(x-y)^2\right )^{1/5}. \]

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