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Find a (real) general solution. State which rule you are using. Show each step of your work. 1.yⁿ + 5y' +4y = 10e⁻³ˣ

User Mcritz
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The given differential equation is a linear homogeneous equation with constant coefficients. The characteristic equation is obtained by replacing y with e^(rt) in the differential equation, which gives:

rⁿe^(rt) + 5re^(rt) + 4e^(rt) = 0

Simplifying the above equation, we get:

rⁿ + 5r + 4 = 0

The roots of the above equation are r = -1, -4. Therefore, the general solution of the differential equation is given by:

y(t) = c₁e^(-t) + c₂e^(-4t) + y_p(t)

where c₁ and c₂ are constants of integration and y_p(t) is a particular solution of the non-homogeneous part of the differential equation.

To find the particular solution, we assume that y_p(t) has the form:

y_p(t) = Ae^(-3t)

where A is a constant to be determined. Substituting this into the differential equation, we get:

Ae^(-3t) - 15Ae^(-3t) + 4Ae^(-3t) = 10e^(-3t)

Simplifying the above equation, we get:

-10Ae^(-3t) = 10e^(-3t)

Therefore, A = -1. Substituting this value into our particular solution, we get:

y_p(t) = -e^(-3t)

Therefore, the general solution of the given differential equation is:

y(t) = c₁e^(-t) + c₂e^(-4t) - e^(-3t)

where c₁ and c₂ are constants of integration.

I hope this helps!

User Dranxo
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