User:Egm4313.s12.team11.sheider/R2.4: K 2011 p.59 pb. 6

Problem Statement
Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given
$$10{y}''-32{y}'+25.6y = 0\!$$

Solution
Initially we modify the original ODE to put it in the form of a second-order homogenous linear ODE with constant coefficients:

$$10{y}''-32{y}'+25.6y = 0\!$$

Dividing both sides by 10:

$${y}''-3.2{y}+2.56y = 0\!$$

The characteristic equation of this ODE is now therefore:

$$\lambda^2 + a\lambda + b = \lambda^2 -3.2\lambda  + 2.56= 0\!$$

Evaluating the discriminant:

$$ a^2 - 4b = (-3.2)^2 - 4(2.56) = 10.24 - 10.24 = 0\!$$

Therefore the equation has a real double root and a general homogenous solution of the form:

$$ y = (c_{1}+c_{2}x)e^{-ax/2}\!$$

And finally we find the general homogenous solution:

$$ y = (c_{1}+c_{2}x)e^{1.6 x}\!$$

Checking:

We found that:

$$ y = (c_{1}+c_{2}x)e^{1.6x}\!$$

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

$$ {y}' = 1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}\!$$

And,

$$ {y}'' = (1.6)^2 c_{1}e^{1.6x}+1.6c_{2}e^{1.6x}+1.6c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}\!$$

$$ {y}'' = (1.6)^2 c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}\!$$

Substituting these equations into the original ODE yields:

$$ 10[(1.6)^2 c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6)^2c_{2}xe^{1.6x}] -32[1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}]+25.6[(c_{1}+c_{2}x)e^{1.6 x}] = 0\!$$

$$ 25.6 c_{1}e^{1.6x}+32c_{2}e^{1.6x}+25.6c_{2}xe^{1.6x} -51.2c_{1}e^{1.6x}-32c_{2}e^{1.6x}-51.2c_{2}xe^{1.6x}+25.6c_{1}e^{1.6 x}+25.6c_{2}xe^{1.6 x} = 0\!$$

$$ (25.6-51.2+25.6) c_{1} e^{1.6x} + (32-32)c_{2}e^{1.6x} + (25.6-51.2+25.6)c_{2}xe^{1.6x} = 0\!$$

$$ 0 \equiv 0 \!$$

Therefore this solution is correct.