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A third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions.y^(3) - 3y'' + 3y' - y = 0; y(0) = 2, y'(0) = 0, y'''(0) = 0; y1 = e^x, y2 = x e^x, y3 = x^2 e^x

Respuesta :

LammettHash

You're given the general solution,

[tex]y=C_1e^x+C_2xe^x+C_3x^2e^x[/tex]

Use the given initial values to find [tex]C_1,C_2,C_3[/tex]:

[tex]y(0)=2\implies2=C_1[/tex]

[tex]y'=C_1e^x+C_2(1+x)e^x+C_3(2x+x^2)e^x[/tex]

[tex]y'(0)=0\implies0=C_1+C_2[/tex]

[tex]\implies C_2=-2[/tex]

[tex]y''=C_1e^x+C_2(2+x)e^x+C_3(2+4x+x^2)e^x[/tex]

[tex]y''(0)=0\implies0=C_1+2C_2+2C_3[/tex]

[tex]\implies C_3=1[/tex]

Then the particular solution is

[tex]y=2e^x-2xe^x+x^2e^x[/tex]

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