Differential Equations (1 Viewer)

faisalabdul16 · Aug 13, 2015

Solve <a href="http://www.codecogs.com/eqnedit.php?latex=x^{2}y''&space;-&space;2xy'&space;+&space;2y&space;=&space;0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^{2}y''&space;-&space;2xy'&space;+&space;2y&space;=&space;0" title="x^{2}y'' - 2xy' + 2y = 0" /></a> by letting <a href="http://www.codecogs.com/eqnedit.php?latex=x&space;=&space;e^{t}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x&space;=&space;e^{t}" title="x = e^{t}" /></a>

InteGrand · Aug 13, 2015

faisalabdul16 said:
Solve <a href="http://www.codecogs.com/eqnedit.php?latex=x^{2}y''&space;-&space;2xy'&space;+&space;2y&space;=&space;0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^{2}y''&space;-&space;2xy'&space;+&space;2y&space;=&space;0" title="x^{2}y'' - 2xy' + 2y = 0" /></a> by letting <a href="http://www.codecogs.com/eqnedit.php?latex=x&space;=&space;e^{t}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x&space;=&space;e^{t}" title="x = e^{t}" /></a>

$x^2 y'' -2xy'+2y=0$

$Let $x=e^t$, then $\frac{\mathrm{d}x}{\mathrm{d}t}=e^t=x,\frac{ \mathrm{d}t}{ \mathrm{d}x}=e^{-t}=\frac{1}{x}$.$

$$\therefore y^\prime =\frac{\mathrm{d}y}{\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x}=\dot{y}e^{-t}\Rightarrow e^t y^\prime = \dot{y}\Rightarrow xy^\prime = \dot{y}$, where we use dots to represent derivatives with respect to $t$.$

$Also, $y'' = \frac{\mathrm{d}}{\mathrm{d}x}\left(y^\prime\right)$$

$=\frac{\mathrm{d}}{\mathrm{d}x} \left( \dot{y}e^{-t}\right)$

$$=\frac{\mathrm{d}}{\mathrm{d}t} \left( \dot{y}e^{-t}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}$ (chain rule)$

$$=\left(\ddot{y}e^{-t}-\dot{y}e^{-t} \right) \times e^{-t}$ (product rule)$

$=\left(\ddot{y}-\dot{y}\right) \times e^{-2t}$

$\Rightarrow y''=\left(\ddot{y}-\dot{y}\right) \times x^{-2}$

$$\Rightarrow x^2 y'' = \ddot{y}-\dot{y}$.$

$Substituting these into the DE, we have$

$\ddot{y}-\dot{y} - 2\dot{y} +2y=0$

$$\Rightarrow \ddot{y}-3\dot{y} +2y=0$.$

$This is now a second order homogeneous ODE with constant coefficients for $y(t)$. Trying $y=e^{\lambda t}$, we obtain the characteristic equation$

$\lambda ^2 -3\lambda+2=0$

$which has roots $\lambda _1 = 1,\lambda_2 =2$.$

$So $y=Ae^{t}+Be^{2t}\Rightarrow y=Ax+Bx^2$, for some arbitrary constants $A$ and $B$.$

faisalabdul16 · Aug 13, 2015

Thank you Integrand!

mreditor16 · Aug 13, 2015

Thanks heaps, Integrand!!

Differential Equations (1 Viewer)

faisalabdul16

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InteGrand

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faisalabdul16

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mreditor16

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