Sample Solutions
for Homework 13(d)(e)(f), 14, 15
Winfried Just
Department of Mathematics, Ohio University
February 10, 2017
Winfried Just, Ohio University
MATH3400, Solution Set 9
Sample Solution for Homework 13(d)
Consider the BVP: y 00 − 10y 0 + 25y = 0, y (0) = 3, y (1) = 2.
1
Find the auxiliary equation: m2 − 10m + 25 = 0.
2
The left-hand side is equal to: (m − 5)2 = 0.
3
Determine the roots: m1 = 5 of multiplicity 2.
4
A fundamental set of solutions: {e 5x , xe 5x }.
5
The general solution is: y (x) = c1 e 5x + c2 xe 5x ,
where c1 , c2 are arbitrary constants.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solution Set 9
Sample Solution for Homework 13(d), completed
Solve the BVP: y 00 − 10y 0 + 25y = 0, y (0) = 3, y (1) = 2.
1
The general of the DE is:
y (x) = c1 e 5x + c2 xe 5x .
2
3
Since the boundary condition does not refer to y 0 (x), we don’t
need to compute it.
Substitute x0 , y0 , y1 from the BC to obtain a system of linear
equations:
y (0) = c1 e 0 + c2 (0)e 0 = c1 = 3,
y (1) = c1 e 5 + c2 (1)e 5 = 2.
4
Substitute c1 = 3 in the second equation and solve for c2 :
We get c2 = 2e −5 − 3 ≈ −2.9865.
5
The solution of the BVP is:
y (x) = 3e 5x + (2e −5 − 3)xe 5x .
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solution Set 9
Sample Solution for Homework 13(e)
Consider the IVP: y 00 − 3y 0 + 10y = 0, y (0) = 0, y 0 (0) = 1.
1
2
3
Find the auxiliary equation: m2 − 3m + 10 = 0.
The left-hand side is equal to: (m − 1.5)2 + 7.75 = 0.
√
√
Determine the roots: m1 = 1.5 + 7.75i, m2 = 1.5 − 7.75i.
4
A fundamental set of solutions:
√
√
{e 1.5x cos 7.75x, e 1.5x sin 7.75x}.
5
The general solution is:
√
√
y (x) = c1 e 1.5x cos 7.75x + c2 e 1.5x sin 7.75x,
where c1 , c2 are arbitrary constants.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solution Set 9
Sample Solution for Homework 13(e), completed
Solve the IVP: y 00 − 3y 0 + 10y = 0, y (0) = 0, y 0 (0) = 1.
1
2
The general of the DE is:
√
√
y (x) = c1 e 1.5x cos 7.75x + c2 e 1.5x sin 7.75x.
Symbolically compute y 0 (x) from the general solution:
√
√
√
y 0 (x) = e 1.5x (1.5c1√cos 7.75x
−
7.75c
sin
7.75x
1
√
√
+1.5c2 sin 7.75x + 7.75c2 cos 7.75x).
3
Substitute x0 , y0 , y1 from the IC:
√
√
y (0) = e 1.5·0 (c1 cos( 7.75 · 0) + c2 sin( 7.75 · 0)) = c1 = 0,
√
y 0 (0) = e 0 (1.5c1 cos 0 + 0 + 0 + 7.75c2 cos 0) = 1,
√
y 0 (0) = 1.5c1 + 7.75c2 = 1.
4
Solve for c1 , c2 : We get c1 = 0, c2 =
5
The solution of the IVP is:
Ohio University – Since 1804
Winfried Just, Ohio University
√1
7.75
≈ 0.3592.
√
y (x) = 0.3592e 1.5x sin 7.75x.
Department of Mathematics
MATH3400, Solution Set 9
Sample Solution for Homework 13(f)
Consider the IVP: y 00 − 7y 0 + 10y = 0, y (0) = 1, y 0 (0) = 3.
1
Find the auxiliary equation: m2 − 7m + 10 = 0.
2
Factor the left-hand side: (m − 2)(m − 5) = 0.
3
Determine the roots: m1 = 2, m2 = 5.
4
A fundamental set of solutions: {e 2x , e 5x }.
5
The general solution is: y (x) = c1 e 2x + c2 e 5x ,
where c1 , c2 are arbitrary constants.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solution Set 9
Sample Solution for Homework 13(f), completed
Solve the IVP:
y 00 − 7y 0 + 10y = 0, y (0) = 1, y 0 (0) = 3.
1
The general solution of the DE is: y (x) = c1 e 2x + c2 e 5x .
2
Symbolically compute y 0 (x) from the general solution:
y 0 (x) = 2c1 e 2x + 5c2 e 5x .
3
Substitute x0 , y0 , y1 from the IC to obtain a system of linear
equations:
y (0) = c1 e 2·0 + c2 e 5·0 = c1 + c2 = 1,
y 0 (0) = 2c1 e 2·0 + 5c2 e 5·0 = 2c1 + 5c2 = 3.
4
Solve for c1 , c2 : We get c1 = 23 , c2 = 31 .
5
The solution of the IVP is:
Ohio University – Since 1804
Winfried Just, Ohio University
y (x) = 23 e 2x + 31 e x .
Department of Mathematics
MATH3400, Solution Set 9
Sample Solutions for Homework 14, 15
Assume a linear homogeneous DE with constant coefficients has a
characteristic polynomial that factors as
Homework 14: (m + 3)4 (m − π)2 (m2 + 6m + 25)3
Its general solution is:
y (x) = e −3x (c1 + c2 x + c3 x 2 + c4 x 3 ) + e πx (c5 + c6 x)
+ e −3x ((c7 + c8 x + c9 x 2 ) cos 4x + (c10 + c11 x + c12 x 2 ) sin 4x)
Homework 15: (m − 4)5 (m2 + 25)(m2 − 2m + 2)3
Its general solution is:
y (x) = e 4x (c1 + c2 x + c3 x 2 + c4 x 3 + c5 x 4 ) + (c6 cos 5x + c7 sin 5x)
+ e x ((c8 + c9 x + c10 x 2 ) cos x + (c11 + c12 x + c13 x 2 ) sin x)
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solution Set 9