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### Messages - Razak Pirani

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##### Quiz 4 / Problem 2 (Day Section)
« on: November 15, 2013, 01:36:33 PM »
7.4 p. 395 #7
Consider the vectors $\mathbf{x}^{(1)}(t) = \begin{pmatrix}t^2\\2t\end{pmatrix}$ and  $\mathbf{x}^{(2)}(t) = \begin{pmatrix}e^t\\e^t\end{pmatrix}$.

(a) Compute the Wronskian of $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$.
$$W(\mathbf{x}^{(1)}(t),\mathbf{x}^{(2)}(t))=t^2 e^t - 2te^t = t(t - 2)e^t.$$

(b) In what intervals are $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$ linearly independent?

When $t=0$ and $t=2$, $W(\mathbf{x}^{(1)}(t),\mathbf{x}^{(2)}(t))$ = 0 and $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$  are linearly dependent.

so $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ and $\mathbf{x}^{(2)}(t)$ and $\mathbf{x}^{(2)}(t)$ are linearly independent at each point except when $t = 0$ and $t = 2$.

(c) What conclusion can be drawn about the coefficients in the system of homogeneous differential equations satisfied by $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$?

If $\mathbf{x}$ satisfies this system $\mathbf{x}'+A\mathbf{x}=0$ then $A$ must be singular at $t=0$ and $t=2$.

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##### Quiz 3 / Problem 1 (Day section)
« on: November 07, 2013, 12:35:09 PM »
4.2 #18 Find the general solution of the given differential equation.

y(6) - y'' = 0

r6 - r2 = 0
r2(r4 - 1) = 0
r2(r2 - 1)(r2 + 1) = 0
r2(r - 1)(r + 1)(r - i)(r + i) = 0

r1,2 = 0
r3 = 1
r4 = -1
r5 = i
r6 = -i

y(t) = c1 + c2t + c3et + c4e-t + c5cost + c6sint

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##### Quiz 1 / Re: Q1, P1--Day section
« on: October 04, 2013, 09:34:54 PM »
Problem 1: Section 2.2, #31

Solve the homogeneous differential equation using the substitution $y(x) = xv(x)$
$$dy/dx = (x^2 + xy + y^2)/x^2$$
First note: Since $y(x) = xv(x)$,  $dy/dx = v(x) + x(dv/dx)$. Dividing the numerator and denominator by $x^2$ and substituting $v = y/x$ yields the homogeneous equation
$$v + x(dv/dx) = 1 + v + v^2\implies dx/x = dv/(1 + v^2).$$
Take the integral of both sides
$$\ln|x| + C = \arctan(v).$$
Substitute $y/x = v$
$$\arctan(y/x) - \ln|x| = C.$$

Observe how I modified the source to provide a proper formatting. Note that the last equation could be resolved with respect to $y$: $y=x\tan (C\ln |x|)$.

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