1st order linear differential equation

• Dec 8th 2007, 10:21 PM
1st order linear differential equation
find the general solution
$\displaystyle 2(2xy+4y - 3)dx + (x+2)^2dy = 0$
• Dec 8th 2007, 11:40 PM
kalagota
Quote:

Originally Posted by ^_^Engineer_Adam^_^
find the general solution
$\displaystyle 2(2xy+4y - 3)dx + (x+2)^2dy = 0$

$\displaystyle \implies \frac{dy}{dx} + \frac{4}{x+2}y = \frac{6}{(x+2)^2}$

Theorem:
the general solution of $\displaystyle \frac{dy}{dx} + q(x)y = r(x)$ is given by $\displaystyle y = cy_1 + y_2$, where $\displaystyle y_1 = \frac{1}{Q(x)}$ and $\displaystyle y_2 = \frac{1}{Q(x)}\int_0^x r(\tau )Q(\tau ) d\tau$ and $\displaystyle Q(x) = \exp \left[ \int_0^x q(\tau ) d\tau \right]$, and also $\displaystyle y_1(0) = 1$, $\displaystyle y_2(0) = 0$ so that $\displaystyle y(0) = c$, the initial condition.

now, from the given, we have $\displaystyle q(x) = \frac{4}{x+2}$ and $\displaystyle r(x) = \frac{6}{(x+2)^2}$

so $\displaystyle Q(x) = \exp \left[ \int_0^x \frac{4}{\tau +2} d\tau \right] = \exp \left[ 4 \left( {\ln (x+2) - \ln 2} \right) \right] = \frac{(x+2)^4}{16}$

therefore, $\displaystyle y_1 = \frac{16}{(x+2)^4}$

$\displaystyle y_2 = \frac{16}{(x+2)^4} \int_0^x \frac{6}{(\tau +2)^2} \cdot \frac{(\tau + 2)^4}{16} \, d\tau = \frac{6}{(x+2)^4} \int_0^x (\tau +2)^2\, d\tau$

$\displaystyle = \frac{6}{(x+2)^4} \left( {\frac{x^3}{3} + 2x^2 + 2x} \right) = \frac{2}{(x+2)^4} \left( {x^3 + 6x^2 + 6x} \right) = \frac{2x^3 + 12x^2 + 12x}{(x+2)^4}$

therefore, the general solution is $\displaystyle y = c\cdot \frac{16}{(x+2)^4} + \frac{2x^3 + 12x^2 + 12x}{(x+2)^4}$
• Dec 9th 2007, 12:12 AM