MATH 225: DIFFERENTIAL EQUATIONS AND LINEAR ALGEBRA ...
MATH 225: DIFFERENTIAL EQUATIONS AND LINEAR ALGEBRA ...
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