Find the general solution of the given second-order differential equation. y'' - 6y' +10y = 0

Solution:

To find the general solution of the equation y'' - 6y' +10y = 0 let us write write down the auxiliary equation for the same.

The auxiliary equation is 

m² - 6m + 10 = 0

The roots of the above equation are calculated from the following formula for the roots of the quadratic equation ax² + bx+ c.

Roots = -b ± √b² - 4ac / 2a

Roots of the equation m² - 6m + 10 = 0 are:

m =  -(-6) ± √(-6)² - 4(1)(10) / 2(1)

=  6 ± √(-6)² - 4(1)(10) / 2

=  6 ± √36 - 40 / 2

=  6 ± √-4 / 2

= 3 ± i

We know that general solution of a second order differential equations with complex roots α ± βi is given by y = e^αx ( C₁ cos(βx) + C₂ sin(βx) ).

Substitute α = 3 and β = 1.

The general solution of the given equation therefore is :

y = e^3x (C₁ cos(x) + C₂ sin(x) )

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Find the general solution of the given second-order differential equation. y'' - 6y' +10y = 0

Summary:

Thus, the general solution of the given second-order differential equation y'' − 6y' + 10y = 0 is y = e^3x (C₁ cos(x) + C₂ sin(x) )