MA1S12: SOLUTIONS TO TUTORIAL 5
1. Solve the initial value problem
√
y 0 = x2 y,
y(0) = 1
Solution: This is a separable first-order ordinary differential equation. We first rewrite the equation as
1
y− 2
dy
= x2
dx
Now we integrate both sides
Z
y
− 12
dy =
Z
1
=⇒ 2y 2 =
x2 dx
x3
+c
3
Using the initial condition
1
2y(0) 2 = 0 + c = 2
The particular solution is
y(x) =
2
x3
+1
6
2. Solve the initial value problem
dy
2
= ex + 2xy,
dx
1
y(0) = 0
2
MA1S12: SOLUTIONS TO TUTORIAL 5
Particular solution for initial value problem in Q.1.
400
300
200
100
-4
-2
2
4
Integral curves for general solution in Q.1.
400
300
200
100
-4
-2
2
4
A slope field for the differential equation in Q.1.
1.0
0.8
0.6
0.4
0.2
0.0
-1.0
-0.5
0.0
0.5
1.0
MA1S12: SOLUTIONS TO TUTORIAL 5
3
Solution: This is a first order linear ordinary differential equation.
First we rewrite the equation in standard form
dy
2
− 2xy = ex
dx
2
Let p(x) = −2x and q(x) = ex . Next we integrate p(x).
Z
p(x) dx =
Z
−2x dx = −x2 + constant
The integrating factor µ(x) is
R
µ(x) = e
p(x) dx
= e−x
2
The general solution to the ODE is
Z
1
y(x) =
µ(x)q(x) dx
µ(x)
Z
2
2
x2
= e
e−x ex dx
Z
x2
= e
1 dx
2
= ex [x + c]
Using the initial condition y(0) = 0 we have
y(0) = e0 [0 + c] = 0 =⇒ c = 0
The particular solution to the ODE is
y(x) = x ex
2
3. Find a general solution to the second-order ODE
y 00 = x2 + 1
4
MA1S12: SOLUTIONS TO TUTORIAL 5
Solution: We reduce this second-order ODE to two first order ODE’s
by making a substitution. Let u(x) = y 0 (x). Then u0 (x) = y 00 (x) =
x2 + 1 which is a separable first order ODE
du
= x2 + 1
dx
Integrating both sides we get
Z
1 du =
Z
=⇒ u =
x2 + 1 dx
x3
+x+c
3
where c is a constant. Next we need to solve the separable first order
ODE
x3
dy
=
+x+c
dx
3
Integrating both sides we get
Z
1 dy =
=⇒ y =
Z
x3
+ x + c dx
3
x4 x2
+
+ cx + d
12
2
where d is a constant. This is the general solution.
4. Find a general solution to the first order ODE
x
dy
= x2 + 3y
dx
x>0
Solution: We first rewrite the equation in standard form
dy
3
− y=x
dx x
MA1S12: SOLUTIONS TO TUTORIAL 5
5
Integral curves for general solution in Q.4.
600
400
200
-4
-2
2
4
-200
-400
-600
We need to find an integrating factor µ(x). Let
p(x) = −
Z
p(x) dx =
Z
3
x
q(x) = x
3
− dx = −3 ln(x) = ln(x−3 ) + const
x
Let
µ(x) = eln(x
−3 )
= x−3
Then the general solution is
Z
1
y(x) =
µ(x)q(x) dx
µ(x)
Z
3
= x
x−2 dx
= x3 −x−1 + c
= cx3 − x2