User:Egm6321.f10.team1.allen/HW2

Given
$$ a_1(x)y'+a_0(x)y=b(x)\ $$

Find
1) Find Solution in terms of $$ a_0\ $$, $$ a_1 \ $$ and $$ b \ $$  for $$ y(x)\ $$ Assuming $$ a_1(x)=1 \ $$ $$ a_0(x)=x \ $$ $$ b(x)=2x+3 \ $$ 2) Find Solution in terms of $$ a_0\ $$ , $$ a_1 \ $$  and $$ b \ $$  for $$ y(x)\ $$ Assuming $$ a_1(x)\ne \ 0\ $$ for all x 3) Find Solution in terms of $$ a_0\ $$, $$ a_1 \ $$ and $$ b \ $$  for $$ y(x)\ $$ Assuming $$ a_1(x)=x^2+1 \ $$ $$ a_0(x)=x \ $$ $$ b(x)=2x \ $$

Solution
$$ Part \ 1) $$

$$ \frac{dy}{dx}+xy=2x+3 $$

The integrating factor is shown and solved for below.

$$ h(x)=\exp\int x \ dx $$

$$ h(x)=\exp(\frac{1}{2}x^2) $$

Multiply both sides by the integrating factor yields

$$ \exp(\frac{1}{2}x^2)\frac{dy}{dx}+\exp(\frac{1}{2}x^2)xy=\exp(\frac{1}{2}x^2)(2x+3) $$

$$ D_x[\exp(\frac{1}{2}x^2)y]=\exp(\frac{1}{2}x^2)(2x+3) $$

$$ \exp(\frac{1}{2}x^2)y=\int\exp(\frac{1}{2}x^2)(2x+3) \ dx $$.

Integration then yields

$$ \exp(\frac{1}{2}x^2)y=2\exp(\frac{1}{2}x^2)+\frac{3}{x}\exp(\frac{1}{2}x^2)+C $$.

Divide both sides by $$ \exp(\frac{1}{2}x^2) $$ gives the final result

$$ y=2+\frac{3}{x}+\frac{C}{\exp(\frac{1}{2}x^2)} $$.

$$ Part \ 3) $$

$$ (x^2+1)\frac{dy}{dx}+xy=2x $$

Divide through by $$ (x^2+1) $$ to obtain

$$ \frac{dy}{dx}+\frac{x}{x^2+1}y=\frac{2x}{x^2+1} $$.

The integrating factor is shown and solved for below.

$$ h(x)=\exp\int\frac{x}{x^2+1}dx $$

$$ h(x)=\exp(\frac{1}{2}ln(x^2+1)) $$

$$ h(x)=(x^2+1)^\frac{1}{2} $$

Multiply both sides by the integrating factor yields

$$ (x^2+1)^\frac{1}{2}\frac{dy}{dx}+xy=2x $$

$$ D_x[(x^2+1)^\frac{1}{2}y]=2x $$

$$ (x^2+1)^\frac{1}{2}y=\int2x \ dx $$

Integration then yields

$$ (x^2+1)^\frac{1}{2}y=x^2+C $$.

Divide both sides by $$ (x^2+1)^\frac{1}{2} $$ gives the final result

$$ y=(x^2+C)(x^2+1)^\frac{-1}{2} $$.

Author

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Given
Eq6, p.10-3 $$ y(x)=\frac{1}{h(x)} \int_{0}^{x} h(s)b(s)\, ds $$ and Eq1, p.10-3 $$ h(x)=e^{ \int_{0}^{x} a_0(s)\, ds } $$

Find
$$ \alpha\ $$ Show that integration factor, $$k_1\ $$, is unnecessary $$ \beta\ $$ Show that Eq6, p.10-3 agrees with King et.al. p.512 i.e. show that $$ y(x)=Ay_h(x)+y_p(x)\ $$ $$ \gamma\ $$ Find $$ y_h(x)\ $$ independantly. i.e. solve $$ y'+a_0y=0\ $$

Solution
$$ \alpha ) $$ If you use $$ \int P(x) \ dx + K_1 $$ instead of $$ \int P(x) \ dx $$,

you will have an integrating factor of $$ h(x)=\exp K_1\exp(\int P(x) \ dx) $$ instead of $$ h(x)=\exp\int P(x) \ dx $$.

If you multiply both sides by the integrating factor you will obtain

$$ \exp K_1 \exp(\int P(x) \ dx)\frac{dy}{dx}+\exp K_1 \exp(\int P(x) \ dx)P(x)y=\exp K_1 \exp(\int P(x) \ dx)Q(x) $$.

As you can see, $$ \exp K_1 $$ will cancel out of the equation rendering it unnecessary.

$$ \cancelto{}{\exp K_1} \exp(\int P(x) \ dx)\frac{dy}{dx}+\cancelto{}{\exp K_1} \exp(\int P(x) \ dx)P(x)y=\cancelto{}{\exp K_1} \exp(\int P(x) \ dx)Q(x) $$.

$$ \beta ) $$ The equation $$ y(x)=\frac{1}{h(x)} \int_{0}^{x} h(s)b(s)\, ds $$ is based off of $$ \frac{dy}{dx}+a_0(x)y=b(x) $$.

It is comparable to $$ y_p=\exp(-\int P(t) \ dt) \int^x Q(x)[\exp(\int^s P(t) \ dt] \ ds $$ which is based off of $$ \frac{dy_p}{dx}+P(x)y_p=Q(x) $$ in K. et al p. 512.

So $$ a_0(x)=P(x) $$ and $$ b(x)=Q(x) $$.

Using the above statements and $$ h(x)=\exp(\int^x P(t) \ dt) $$ yields $$ y_p=\frac{1}{h(x)} \int_{0}^{x} h(s)b(s)\, ds $$.

$$ \gamma ) $$ Since $$ \exp(x) $$ is repeating after integration, it is ideal for this solution.

Through an educated guess, $$ y=C\exp(-a_0x) $$ where C is any arbitrary constant.

Egm6321.f10.team1.allen 03:45, 22 September 2010 (UTC)