User:Egm6321.f09.Team1.sallstrom/HW2

= Problem on page 7-1 =

= Problems on pages 8-3 and 8-4 = Given the equation

y' + \frac{1}{x} y = x^2 $$
 * 1) Show that the solution for the integrating factor is $$h(x)=x$$
 * 2) Solve for y

Solution
Using the form

h y' + h' y = (hy)' = h b $$ and substituting

h(x) = e^{g(x)} $$ We have that

h y' + h' y = h \left( y' + g' y \right) = h \left( y' + \frac{1}{x} y \right) $$ So

g'(x) = \frac{1}{x} \quad \Rightarrow g(x) = \ln(x) + C $$ We can assume that $$C=0$$ without changing the end result. This yields

h(x) = e^{ln(x)} = x $$ Now, integrating

(hy)' = h b $$ yields the final solution

y(x) = \frac{1}{h(x)} \int^x h(s) b(s) ds = \frac{1}{x} \int^x s^3 ds = \frac{1}{x} \left[ \frac{x^4}{4} + C \right] = \frac{x^3}{4} + \frac{C}{x} $$ where $$C$$ is a constant.

= Problem on page 9-2 = With the definition

\bar b(x) := \int^x b(s) ds $$ and the equation

\bar b(x) y'(x) + \left[a(x) y(x) + k \right] = 0 $$ Let
 * $$ a(x) = x^4 \frac{}{}$$
 * $$ b(x) = x \Rightarrow \bar b(x) = \frac{1}{2} x^2 $$
 * $$ k = 10 \frac{}{} $$

This yields the equation

\frac{1}{2} x^2 y' + \left[ x^4 y + 10 \right] = 0 $$ Show that the above equation has an exact solution

Solution
Rewrite the given equation

y' + 2 x^2 y = -20 x^{-2} $$ Then multiply by $$h(x) = e^{g(x)}$$

h(x) y' + h(x) 2 x^2 y = -20 h(x) x^{-2} $$ Note that $$h'(x) = g'(x) h(x)$$, and substitute

h(x) y' + \frac{h'(x)}{g'(x)}2 x^2 y = -20 h(x) x^{-2} $$ Now, we use

g'(x) = 2 x^2 \Rightarrow g(x) = \frac{2}{3} x^3 $$ So that

h(x) y' + h'(x) y = -20 h(x) x^{-2} $$

\left( h(x) y \right)'= -20 h(x) x^{-2} $$ Now, integrate both sides

y = -\frac{1}{h(x)} \int^x 20 h(s) s^{-2} ds $$ Substitute in $$h(x)$$

y = -\frac{1}{\exp(2 x^3/3)} \int^x 20 \exp(2 s^3/3) s^{-2} ds $$