Question

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section 4.2, y2 = y1(x) e−∫p(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). 9y'' − 12y' + 4y = 0; y1 = e2x/3

Answer 1

Given that $y_1=e^{2x/3}$, we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$.

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$, so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

Answer 2

The indicated function y1(x) is a solution of the given differential equation.The general solution is $y = c_1 e^{2x}- c_2e^{-6x}/8$

What is a differential equation?

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.

The derivatives might be of any order, some terms might contain the product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

Given differential equation is

y''-4y'+4y=0

and

$y_1(x) = e^{2x}$

$y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx$

The general form of equation

y''+P(x)y'+Q(x)y=0

Comparing both the equation

So, P(x)= - 4

$y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx$

$y_2(x) = e^{2x}\int e^{-4x} \, / e^{4x}dx$

$y_2(x) = e^{2x}\int e^{-8x}dx\\\\y_2(x) = -e^{-6x}/8$

The general solution is

$y = c_1 e^{2x}- c_2e^{-6x}/8$

Learn more about differential equations;

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