Question

(y^{4})-16y=(e^2x)-(15cos(x))

ojo.... no es potencia, es la 4 derivada, es una EDO lineal homogénea con coeficientes constantes....

Answer 1

Differential Equations

Solution:

The given differential equation is

$\quad (D^{4}-16)\:y=e^{2x}-15\:cosx$

To find C.F.

Auxiliary equation is

$\quad m^{4}-16=0$

$\Rightarrow m=2,\:-2,\:\pm 2i$

$\therefore$ C.F. $=C_{1}e^{2x}+C_{2}e^{-2x}+C_{3}\:cos2x+C_{4}\:sin2x$

To find P.I.

We assume a particular solution $y_{p}$ of the form

$\quad y_{p}=Axe^{2x}+B\:sinx+C\:cosx$

$\Rightarrow Dy_{p}=2Axe^{2x}+Ae^{2x}+B\:cosx-C\:sinx$

$\Rightarrow D^{2}y_{p}=4Axe^{2x}+4Ae^{2x}-B\:sinx-C\:cosx$

$\Rightarrow D^{3}y_{p}=8Axe^{2x}+12Ae^{2x}-B\:sinx-C\:cosx$

$\Rightarrow D^{4}y_{p}=16Axe^{2x}+32Ae^{2x}+B\:sinx+C\:cosx$

We have

$\quad (D^{4}-16)\:y_{p}=e^{2x}-15\:cosx$

$\Rightarrow 16Axe^{2x}+32Ae^{2x}+B\:sinx+C\:cosx-16Axe^{2x}-16B\:sinx-16C\:cosx=e^{2x}-15\:cosx$

$\Rightarrow 32Ae^{2x}-15B\:sinx-15C\:cosx=e^{2x}-15\:cosx$

Comparing among coefficients, we get

$\quad A=\frac{1}{32},\:B=0,\:C=1$

So we have

$\quad y_{p}=\frac{1}{32}xe^{2x}+cosx$

Finding complete solution.

Therefore the general solution is

$y=C_{1}e^{2x}+C_{2}e^{-2x}+C_{3}\:cos2x+C_{4}\:sin2x+\frac{1}{32}xe^{2x}+cosx$

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