Download MAS110 Solutions: Differential equations

MAS110 Solutions: Differential equations
1.
R
R
dy
3x2
=
=⇒ 2y dy = 3x2 dx =⇒ y 2 = x3 + C.
dx
2y
Also y(0) = 2 =⇒ C = 4; so y 2 = x3 + 4. Therefore y =
have taken positive root since y(0) = 2.
2.
√
x3 + 4 and we
p
R
R
dy
√ 1 dy = 2x dx =⇒ arcsin y = x2 + C =⇒
= 2x 1 − y 2 =⇒
2
1−y
dx
y = sin(x2 + C).
3. Sub y = ux. As
dy
dx
=
3
du
dx
3
ux + x
x + u, we obtain
du
+ u = u3 x3 + 2u2 x3 − x3
x
dx
which simplifies to
x(u + 1)
du
= u3 + u2 − u − 1 = (u + 1)(u2 − 1).
dx
Thus either u + 1 = 0 which gives y = −x, which is indeed a solution, or
du
x
= u2 − 1 in which case we obtain
dx
du
x
= u2 − 1
dx
Z
Z
1
1
=⇒
du =
dx
u2 − 1
x
Z
Z 1
1
1
1 =⇒
−
du =
dx
2 u−1 u+1
x
1 u − 1 =⇒ ln = ln |x| + C
2
u + 1
u − 1
= C 0 x2
(new constant)
=⇒ u + 1
y−x
=⇒
= Ax2
(new constant)
y+x
=⇒ y − x = Ax2 (y + x)
=⇒ y − Ayx2 = x + Ax3
=⇒ y =
x(1 + Ax2 )
.
1 − Ax2
(Note that the solution y = −x appears as the limiting case when A → ∞.)
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4.
(i)
ex y 0 + 2ex y = 1 =⇒
=⇒
e2x y 0 + 2e2x y = ex
(mulitply by ex )
d 2x e y = ex
dx
Z
Z
2x
d e y = ex dx
=⇒
e2x y = ex + C
=⇒
=⇒ y = e−x + Ce−2x .
(ii)
y 0 + xy = x =⇒
x2
x2
x2
=⇒
e 2 y 0 + xe 2 y = xe 2
x2
d x2 e 2 y = xe 2
Z
Zdx
x2
x2 d e 2 y = xe 2 dx
=⇒
e
=⇒
x2
2
y=e
x2
2
+C
(mulitply by e
=⇒ y = 1 + Ce−
x2
2
x2
2
)
.
(iii)
x3 y 0 + 5x2 y =
sin x
=⇒
x2
x5 y 0 + 5x4 y = sin x
=⇒
d 5 x y = sin x
Zdx
Z
5
d x y = sin x dx
=⇒
x5 y = − cos x + C
=⇒
(mulitply by x2 )
=⇒ y = −
cos x
C
+ 5.
5
x
x
(iv)
(tan x)y 0 + y = sin3 x =⇒
=⇒
=⇒
(sin x)y 0 + (cos x)y = sin3 x cos x
d
(sin x)y = sin3 x cos x
Zdx
Z
d (sin x)y = sin3 x cos x dx
(mulitply by cos x)
1
1
C
sin4 x + C
=⇒ y = sin3 x +
.
4
4
sin x
√
−2 ± −4
2
5. The auxiliary equation λ + 2λ + 2 = 0 has roots
= −1 ± i. Hence
2
the general solution is y = e−t A cos t + B sin t .
Now y(0) = 1 implies 1 = A. So y = e−t cos t + B sin t . Also,
y 0 = e−t − sin t + B cos t − e−t cos t + B sin t and y 0 (0) = −1
=⇒
y sin x =
implies −1 = B − 1 i.e. B = 0. So the required solution is y = e−t cos t.
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6. The auxiliary equation λ2 + 2λ − 3 = 0 has roots 1 and −3. Hence the general
solution is y = Aet + Be−3t . Now y 0 = Aet − 3Be−3t . So y(0) = 1 and
y 0 (0) = −1 imply A + B = 1 and A − 3B = −1, and so A = B = 21 . Therefore
the required solution is y = 12 et + 12 e−3t .
7. The auxiliary equation λ2 + 2λ + 1 = 0 has repeated roots −1 and −1. Hence
the general solution is y = (A + Bt)e−t .
8. From the preceding question the homogeneous part has general solution yh =
(A + Bt)et .
Try yp = a cos t + b sin t as a particular solution. yp0 = −a sin t + b cos t and
yp00 = −a cos t − b sin t, so
yp00 + 2yp + yp = (−a − 2a + a) cos t + (−b + 2b + b) sin t = −2a cos t + 2b sin t.
Thus a = 0 and b = 21 , and yp =
1
2
sin t.
The required general solution therefore is y = yp + yh =
1
2
sin t + (A + Bt)e−t .
9. The homogeneous part has general solution yh = (A + Bt)e−t .
Try yp = aet as a particular solution. yp00 + 2yp + yp = 4aet , so a = 41 .
The required general solution therefore is y = 14 et + (A + Bt)e−t .
10. The homogeneous part has general solution yh = (A + Bt)e−t .
Try yp = at2 e−t as a particular solution. yp0 = −at2 e−t + 2ate−t and yp00 =
at2 e−t − 2ate−t − 2ate−t + 2ae−t = at2 e−t − 4ate−t + 2ae−t , so
yp00 + 2yp + yp = at2 − 4at + 2a − 2at2 + 4at + at2 e−t = 2ae−t
and a = 21 .
The required general solution therefore is y = 21 t2 e−t + (A + Bt)e−t .
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