Additional problems, week 1-4
Example 1. Compute the solution of the pde
yzx−xzy = 0.
Example 2. Compute the solution of the pde
zx+yzy = 0.
Example 3. Compute the solution of the pde
√1−x2zx+zy = 0, z(0, y) =y.
Example 4. Find the integral surfaces of the vector field (x2, y2,(x+y)z)T containing the line
(a) x=y =z, (b) x= 1, y=z, (c) x=y,z =x2.
Example 5. Compute the solution of the pde zux+yuy+xuz = 0.
Example 6. Compute the solution of the pde x2ux+y2uy+ (x+y)zuz = 0.
Example 7. Solve the pdes (a) azx+bzy =−cz, (b) xzx+ (y−1)zy =xz.
Example 8. Solve the initial value problem
xzx+ (y−1)zy =xz z(x,0) =g(x).
Example 9. Find the general solution u(x, y, z) of the equation
Solutions for the additional problems
Solution 1. Here P =y, Q=−x and we get dy
dx = Q P dx
y =− dy x xdx=−ydy x2 =−y2+C
u(x, y) =C =x2+y2
z(x, y) =f(u(x, y)) =f(x2+y2).
Solution 2. With P = 1 and Q=y we get dx
1 = dy y x= ln|y|+C
u(x, y) =C =x−ln|y|
z(x, y) =f(u(x, y)) =f(x−ln|y|).
Solution 3. We get
√ dx
1−x2 = dy 1 arcsin(x) =y+C
u(x, y) =C = arcsin(x)−y
z(x, y) =f(u(x, y)) =f(arcsin(x)−y).
The initial condition leads to
f(arcsin(0)−y) =y f(−y) =y
z(x, y) =y−arcsin(x).
Solution 4. The equations read dx
x2 = dy
y2 = dz (x+y)z and first we use
dx x2 = dy
y2
− 1
x =−1 y +c c1 =u1(x, y, z) = 1
x − 1 y. Next we use
dx
x2 = dz (x+y)z 1
x + y x2
dx= dz
z |y= 1
x−c −1
1
x + 1
x−c1x2
dx= dz z
ln|x|+ ln|x−c1x2| −2 ln|1−c1x|= ln|z|+c x(x−c1x2)
(1−c1x)2 =c2z x(x−x+ xy2)
(1−1 + xy)2 =c2z x3/y
(x/y)2 =c2z
xy =c2z
c2 =u2(x, y, z) = xy z .
Solution for (a): It holdsϕ(t) =ψ(t) =χ(t) and we get
1 1
This results in
f(0, ϕ(t)) = 0 f(F1, F2) = F1
f 1
x − 1 y,xy
z
= 1 x − 1
y = 0 1
x− 1 y = 0.
Solution for (b): Here c= (1, ψ(t), ψ(t) results in u1 = 1− 1
ψ(t), u2 = 1 f(1−ψ(t)−1,1) = 0 f(F1, F2) = F2−1 f
1 x − 1
y,xy z
= xy
z −1 = 0 xy
z = 1 xy =z.
Solution for (c): Here we havec= (ϕ(t), ϕ(t), ϕ(t)2)T and this results in u1 = 1
ϕ(t) − 1 ϕ(t) = 0 u2 = ϕ(t)ϕ(t)
ϕ(t)2 = 1 f(0,1) = 0
f(F1, F2) = g(F1)−F2 withg(0) = 1 g
1 x − 1
y
−xy z = 0 zg
1 x − 1
y
=xy withg(0) = 1.
Solution 5. The solution of the pdezux+yuy +xuz = 0 can be computed as follows dx
z = dy y = dz
x . First we use
dx z = dz
x xdx=zdz
x2 2 = z2
2 −˜c
c1 =u1(x, y, z) =x2 −z2. Next we use
dx z = dy
y
√ dx x2 −c1
= dy y ln|x+p
x2−c1|= ln|y|+ ˜c x+p
x2 −c1 =c2y x+z=c2y
c2 =u2(x, y, z) = x+z y and the final solution reads
u(x, y, z) =f(u1, u2) =f(x2−z2,x+z y ).
Solution 6. From solution 4 we know
c1 =u1(x, y, z) = 1 x − 1
y, c2 =u2(x, y, z) = xy
z . Finally we get
Solution 7. Solution for (a): In order to compute the integral curves we get dx
a = dy
b = dz
−cz. First we solve
bdx=ady bx=ay+c1
c1 =u1(x, y, z) =bx−ay.
Next we get dx
a = dz
−cz
− c
a dx= dz z
− c
ax= ln|z˜c exp
−c ax
= z c2
c2 =u2(x, y, z) =zexpc ax
. So we get the form
u2 =ze
c ax
=f(bx−ay) z =f(bx−ay)e−
c ax.
Solution for (b): The integral curves are computed for dx
x = dy
y−1 = dz xz. Integral curves:
ln|x|= ln|y−1|+ ˜c x= (y−1)c1
c1 =u1(x, y, z) = x y−1
and dx
x = dz xz dx= dz z x= ln|z| −˜c ex = z
c2
c2 =u2(x, y, z) =ze−x. This results in
ze−x =f(x/(y−1)) z =exf(x/(y−1)).
Solution 8. The general solution reads
z =exf(x/(y−1)).
With the initial condition we get
z(x,0) =exf(x/(0−1)) =exf(−x) =g(x) f(−x) =g(x)e−x ⇒f(t) =g(−t)et z(x, y) =exg
x 1−y
e
x y−1
=e
x2
y−1g(x/(1−y)). Solution 9. Integral curves:
dx
x = dy
y−1 = dz xz, c1 =u1(x, y, z) = x
y−1, c2 =u2(x, y, z) =ze−x.