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MATH 225: DIFFERENTIAL EQUATIONS AND LINEAR ALGEBRA ...

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2.b) What conditions must b1,b2, and b3 satisfy in order for the system<br />

x1 + x2 + 2x3 = b1,<br />

x1<br />

to be consistent. Find the solutions.<br />

+ x3 = b2,<br />

2x1 + x2 + 3x3 = b3<br />

Solution :<br />

⎡<br />

1<br />

⎣1<br />

1<br />

0<br />

2<br />

1<br />

⎤<br />

b1<br />

b2⎦<br />

⎡<br />

1<br />

→ ⎣0<br />

1<br />

−1<br />

2<br />

−1<br />

⎤<br />

b1<br />

b2 − b1⎦<br />

⎡<br />

1<br />

→ ⎣0<br />

1<br />

−1<br />

2<br />

−1<br />

⎤<br />

b1<br />

b2 − b1 ⎦<br />

2 1 3 b3 2 1 3 b3 0 −1 −1 b3 − 2b1<br />

⎡<br />

1<br />

→ ⎣0<br />

1<br />

1<br />

2<br />

1<br />

b1<br />

b1 − b2<br />

⎤<br />

⎦<br />

0 0 0 b1 − b3 + b2<br />

Hence for consistency b3 = b1 + b2. In this case there are infinitely many solutions<br />

x1 = b2 − t, x2 = b1 − b2 + t, x3 = t where t ∈ R is an arbitrary real variable.<br />

3.a) Solve the differential equation x 2 y ′ + 2xy = 5y 4 with y(1) = 1. Discus the<br />

existence and uniqueness of the solution of the differential equation with the initial<br />

condition y(0) = 0.<br />

Solution : Let v(x) = y −3 then<br />

v ′ − 6<br />

x<br />

15<br />

v + = 0<br />

x2 This is a linear equation with the integrating factor ρ(x) = 1<br />

x 6 . Hence the solution<br />

is<br />

v(x) = C x 6 + 15<br />

7<br />

1<br />

x<br />

�<br />

5x<br />

→ y(x) =<br />

7Cx7 + 15<br />

where C is an arbitrary constant. If y(1) = 1 then C = − 8<br />

7 .<br />

The singular solution of this equation is y = 0. Furthermore at x = 0 the existence<br />

and uniqueness is not guaranteed. It is clear from the solution that y(0) = 0 is<br />

satisfied for all C, hence there are infinitely many solutions.<br />

3.b) Solve the differential equation (x + ln y)dx + ( x<br />

y + ey )dy = 0 with y(0) = 1<br />

Solution: M(x, y) = x + ln y and N(x, y) = x<br />

y + ey . Then<br />

but<br />

Fx = M(x, y) = x + ln y → F (x, y) = x2<br />

2<br />

� 1/3<br />

+ x ln y + g(y),<br />

Fy = x<br />

y + g′ = x<br />

y + ey , → F (x, y) = x2<br />

+ x ln y + ey<br />

2

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