3 years ago

Please help ZB= Round your answer to the nearest hundredth. B. 6 5 с

Mathematics

1 answer:

Aloiza [94]3 years ago

8 0

Bhis the answer u can write

You might be interested in

Differential Equations - Reduction of order

iren [92.7K]

With reduction of order, we assume a solution of the form $y_2=zy_1=ze^{2x}$ , with $z=z(x)$ . Then

${y_2}'=(z'+2z)e^{2x}$
${y_2}''=(z''+4z'+4z)e^{2x}$

and substituting into the ODE gives

$x(z''+4z'+4z)e^{2x}-(2x+1)(z'+2z)e^{2x}+2ze^{2x}=0$
$x(z''+4z'+4z)-(2x+1)(z'+2z)+2z=0$
$xz''+(2x-1)z'=0$

Let $\xi(x)=z'(x)$ , so that $\xi'=z''$ . This gives the linear ODE

$x\xi'+(2x-1)\xi=0$

This equation is also separable, so you can write

$\dfrac{\xi'}{\xi}=\dfrac{1-2x}x$

Integrating both sides with respect to $x$ gives

$\ln|\xi|=-2x+\ln x+C_1$
$\xi=C_1xe^{-2x}$

Next, solve $z'=\xi$ for $z$ by integrating both sides again with respect to $x$ .

$z'=\xi$
$\implies z=\displaystyle\int C_1xe^{-2x}\,\mathrm dx$
$\implies z=C_1e^{-2x}(2x+1)+C_2$

And finally, solve for $y_2$ .

$y_2=zy_1=C_1(2x+1)+C_2e^{2x}$

and note that $y_1$ is already taken into account as part of $y_2$ , so this is the general solution to the ODE.

6 0

3 years ago

If necessary, round your answer to the nearest tenth,

Inessa [10]

Check the picture below.