# MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Wronskian, The Roots

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Solutions of Test 2

Jordan Bell

July 9, 2013

1. The characteristic equation is r2+3r−10 = 0, and solving gives r=−5,2.

So solutions of the homogeneous equation are y1=e−5tand y2=e2t. Calculate

the Wronskian: we get W= 7e−3t.

To ﬁnd u1we’ll end up doing integration by parts (going from third to fourth

line).

u1=−Zy2g

Wdt

=−Ze2t7tet

7e−3tdt

=−Zte6tdt

=−te6t

6+e6t

36 .

To ﬁnd u2we’ll end up doing integration by parts (going from third to fourth

line).

u2=Zy1g

Wdt

=Ze−5t7tet

7e−3tdt

=Zte−tdt

=−te−t−e−t.

Thus we can write the solution of the initial value problem as

y=c1e−5t+c2e2t+e−5t−te6t

6+e6t

36 +e2t−te−t−e−t

=c1e−5t+c2e2t−7

6tet−35

36et.

If we want to choose c1and c2to satisfy the initial conditions, we’ll need to

evaluate y(0) and y0(0).

1

y(0) = c1+c2−35

36.

y0(t) = −5c1e−5t+ 2c2e2t−7

6et−7

6tet−35

36et

y0(0) = −5c1+ 2c2−7

6−35

36

Using the initial conditions we get c1=−2

7and c2=2

7. Thus the solution

of the initial value problem is

y(t) = −2

7e−5t+2

7e2t−7

6tet−35

36et.

2. The ﬁrst equation gives

u0

1=−y2

y1

u0

2.

Putting this into the second equation gives

−y0

1

y2

y1

u0

2+y0

2u0

2=g.

Multiplying by y1gives

−y0

1y2u0

2+y1y0

2u0

2=gy1,

i.e.,

u0

2(y1y0

2−y0

1y2) = gy1,

or,

u0

2W=gy1.

Thus

u2=Zgy1

Wdt.

Since u0

2W=gy1, using u0

1=−y2

y1u0

2gives us

u0

1=−y2

y1

gy1

W=−gy2

W,

and thus

u1=−Zgy2

Wdt.

3. Use the fact that y00

2=−py0

2−qy2and y00

1=−py0

1−qy1. Then

−y1y00

2−y00

1y2

W=−y1(−py0

2−qy2)−(−py0

1−qy1)y2

W

=−−py1y0

2−qy1y2+py0

1y2+qy1y2

W=py1y0

2−py0

1y2

W=p.

2