Respuesta :

Answer:

[tex]y_p=A+Bt+Ce^{2t}[/tex]

Step-by-step explanation:

Given: [tex]y'' - 2y' = 6t + 5e^{2t}[/tex].

we need to find the correct form for [tex]y_p[/tex] if the equation is solve using undetermined coefficients.

A first order differential equation [tex]\frac{\mathrm{d} y}{\mathrm{d} x}=f\left ( x,y \right )[/tex] is said to be homogeneous if [tex]f(tx,ty)=f(x,y)[/tex] for all t.

Consider homogeneous equation [tex]y''-2y'=0[/tex]

Let [tex]y=e^{rt}[/tex] be the solution .

We get [tex](r^2-2r)e^{rt}=0[/tex]

Since [tex]e^{rt}\neq 0[/tex], [tex]r^2-2r=0[/tex].

So, we get solution as [tex]y_c=c_1+c_2e^{2t}[/tex]

As constant term and [tex]e^{2t}[/tex] are already in the R.H.S of equation

[tex]y" - 2y' = 6t + 5e^{2t}[/tex], we can take [tex]y_p[/tex] as [tex]y_p=A+Bt+Ce^{2t}[/tex]

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