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anonymous

  • 5 years ago

Find y as a function of x if y^(4)−4y'''+4y''=128e^(−2x) y(0)=9 y'(0)=6 y''(0)=12 y'''(0)=−16

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  1. anonymous
    • 5 years ago
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    This is a nonhomogeneous differential equation with constant equation. It has a solution of the form: \[y(x)=y_c+y_p\] where y_c is the complementary solution associated with its homogeneous equation, and y_p is the particular solution.

  2. anonymous
    • 5 years ago
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    To find y_c, we just have to solve the auxiliary equation associated with the homogeneous equation, that's: \[m^4-4m^3+4m^2=0 \implies m^2(m-2)^2=0\] Which has double roots of 0 and 2. Then: \[y_c=c_1+c_2x+c_3e^{2x}+c_4xe^{2x}\]

  3. anonymous
    • 5 years ago
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    Now, let's assume a particular solution Ae^(-2x), substitute in the original differential equation: \[16Ae^{-2x}-4(-8Ae^{-2x})+4(4Ae^{-2x})=128e^{-2x} \implies 16A+36A+16A=128\] and hence A=2, therefore the particular solution is: \[y_p=2e^{-2x}\]

  4. anonymous
    • 5 years ago
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    Therefore the general solution is: \[y=c_1+c_2x+c_3e^{2x}+c_4xe^{2x}+2e^{-2x}\] Now, use the given initial conditions to find the constants c_1, c_2,c_3 and c_4.

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