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GenaCL600 [577]
3 years ago
9

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

nts. If possible, factor L. (Used for the differential operator.) 3y'' − 8y' − 3y = 4
Mathematics
1 answer:
melamori03 [73]3 years ago
4 0

Answer:

The complete solution is

\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

y = e^{mx}

Differentiating with respect to x

y'= me^{mx}

Again differentiating with respect to x

y''= m ^2 e^{mx}

Putting the value of y, y' and y'' in left side of the differential equation

3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0

\Rightarrow 3m^2-8m-3=0

The auxiliary equation is

3m^2-8m-3=0

\Rightarrow 3m^2 -9m+m-3m=0

\Rightarrow 3m(m-3)+1(m-3)=0

\Rightarrow (3m+1)(m-3)=0

\Rightarrow m = 3, -\frac13

The complementary function is

y= Ae^{3x}+Be^{-\frac13 x}

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)    

For particular integral

y_p=\frac 1{(3D+1)(D-3)} .4

    =4.\frac 1{(3D+1)(D-3)} .e^{0.x}    [since e^{0.x}=1]

   =4\frac{1}{(3.0+1)(0-3)}      [ replace D by 0 , since L(0)≠0]

   =-\frac43

The complete solution is

y= C.F+P.I

\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43

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