2nd order linear differential equation (homogeneous)

In summary, the conversation discussed solving the equation 354y`` −692y` + 235y =0 with initial conditions y(0) = 7 and y`(0) = 4. The solution involved dividing the equation by 354, finding the roots, and using the initial conditions to solve for the constants. The incorrect answer was due to careless rounding, while using fractions instead of rounded numbers gave the correct solution.
  • #1
jumbogala
423
4

Homework Statement


Solve 354y`` −692y` + 235y =0

y(0) = 7
y`(0) = 4

Homework Equations


The Attempt at a Solution


First I divided the equation by 354 to get y`` - 1.56y` + 0.894y = 0.

Then I found the roots of this to be 0.94, repeated twice.

For repeated roots the solution looks like y= C1e0.94t + C2te0.94t

Using the initial conditions, solve. You find that C1 = 7.

y` = 0.94C1e0.94t + 0.94C2te0.94t + C2e0.94t. Plugging in t = 0 we find that C2 = -2.61.

Therefore y = y= 7e0.94t -2.61te0.94t

But this isn't the correct answer. Where did I go wrong?
 
Last edited:
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  • #2
You've rounded off incorrectly. -612/324 is 1.89 to two places, not 1.88. But I would be inclined to leave it as a fraction: 17/9. Are you allowed to do that?
 
  • #3
Apart from the fact that you rounded nearly every single number, why isn't the solution correct?
Because I got the same (up to rounding errors).
 
  • #4
Okay, I used fractions in my answer and that worked. Next time I'll be careful not to round carelessly... thanks!
 

What is a 2nd order linear differential equation?

A 2nd order linear differential equation is a mathematical equation that involves a function, its derivatives, and constants. It can be written in the form of y'' + p(t)y' + q(t)y = 0, where y is the dependent variable and p(t) and q(t) are functions of the independent variable t.

What does it mean for a 2nd order linear differential equation to be homogeneous?

A 2nd order linear differential equation is considered homogeneous if all terms in the equation have the same degree of the dependent variable and its derivatives. In other words, there are no constants or terms without derivatives in the equation.

How do you solve a 2nd order linear differential equation (homogeneous)?

The general solution to a 2nd order linear differential equation can be found by using the method of undetermined coefficients or the method of variation of parameters. Both methods involve finding a particular solution and adding it to the complementary function, which is the solution to the corresponding homogeneous equation.

What is the characteristic equation for a 2nd order linear differential equation (homogeneous)?

The characteristic equation for a 2nd order linear differential equation is found by substituting y = e^(rt) into the equation and solving for r. This results in a quadratic equation, known as the characteristic equation, which is used to find the roots or values of r.

What is the role of initial conditions in solving a 2nd order linear differential equation (homogeneous)?

Initial conditions, also known as boundary conditions, are required when solving a 2nd order linear differential equation to find a specific solution. These conditions specify the values of the dependent variable and its derivatives at a certain point or points in the domain, allowing for the determination of unique values for the constants in the general solution.

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