solve (D³+2D²+D)y=e^2x+x²+x
Answers
Step-by-step explanation:
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Answer:
The general equation is y(x) = c₁ (e⁰ˣ) + c₂ (e⁻ˣ) + c₃ (e⁻ˣ) + 1/2x (e²ˣ)
Step-by-step explanation:
To solve the differential equation (D³+2D²+D) y = e²ˣ + x² + x, we can first divide both sides by D³+2D²+D:
y = ( e²ˣ + x² + x ) / ( D³ + 2D² + D )
This equation is in the form of a "variation of parameters" and can be solved by finding a particular solution and a complementary solution.
Firstly, we need to find the complementary solution which is the general solution for the homogeneous equation D³y+2D²y+Dy=0
The characteristic equation is r³ + 2r² + r = 0, which has three roots r₁ = 0, r₂ = (-1) , r₃ = (-1).
Thus, the complementary solution is yc(x) = c₁ (e⁰ˣ) + c₂ (e⁻ˣ) + c₃ (e⁻ˣ)
Now we need to find a particular solution for the non-homogeneous equation: D³y + 2D²y + Dy = e²ˣ + x² + x
A possible method is to try a solution of the form y p(x) = Axe²ˣ
Substituting it into the non-homogeneous equation, we get:
D³(Axe²ˣ) + 2D²(Axe²ˣ) + D(Axe²ˣ) = e²ˣ + x² + x
Expanding, we get:
2Ae²ˣ + (6A+2A)x + (2A+2A) = e²ˣ + x² + x
Equating coefficients, we get:
2A = 1, 6A+2A = 0, 2A+2A = 0
Solving for A, we get A = 1/2
Therefore, the particular solution is yp(x) = 1/2xe²ˣ
Finally, the general solution is y(x) = yc(x) + yp(x) = c₁ (e⁰ˣ) + c₂ (e⁻ˣ) + c₃ (e⁻ˣ) + 1/2x (e²ˣ)
Note: c₁, c₂, c₃ are arbitrary constants that can be determined by using initial/boundary conditions.
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