Why Is y''(a) Determined by y(a) and y'(a) in a Differential Equation?

[AndreaA]

Indicate why we can impose only n initial conditions on a solution of nth order linear differential equation.

A) Given the equation y'' + py'+ qy = 0
explain why the value of y''(a) is determined by the values of y(a) and y'(a).

B) Prove that the equation y'' - 2y' -5y =0
has the solution satisfying the conditions y(0) = 1, y'(0) = 0, and y''(0) = C
if and only if C = 5.

 

[matematikawan]

A) Given the equation y'' + py'+ qy = 0
explain why the value of y''(a) is determined by the values of y(a) and y'(a).

The DE is linear so it must have a general solution which is a linear combination of two linearly independent solutions y₁(x) and y₂(x).

y(x)=c₁y₁(x) + c₂y₂(x) .

c₁ and c₂ can be determined uniquely from the given initial conditions. So the result can be deduce from here.

 

[deluks917]

Do you need to the dimension of the solution set to a second order system is two dimensional?

 

[AlephZero]

A) Given the equation y'' + py'+ qy = 0
explain why the value of y''(a) is determined by the values of y(a) and y'(a).

Because y''(a) = -py'(a) - qy(a).

Move along, please, there's nothing to explain here...

 

[blue_raver22]

A) In this equation, y'' represents the second derivative of y with respect to the independent variable, while y' represents the first derivative of y. This means that the value of y'' at a specific point a is determined by the rate of change of y' at that point, which in turn is determined by the value of y at that point. This is because the second derivative measures the curvature or concavity of a function, and this curvature is influenced by the slope or steepness of the first derivative, which is in turn influenced by the value of the function itself. Therefore, the value of y'' at a is directly related to the values of y(a) and y'(a).

B) To prove this statement, we can start by substituting the given initial conditions into the equation y'' - 2y' -5y =0 and solving for C.

Substituting y(0) = 1, y'(0) = 0, and y''(0) = C into the equation, we get:
y''(0) - 2y'(0) -5y(0) = 0
C - 2(0) - 5(1) = 0
C - 5 = 0
C = 5

This shows that the only value of C that satisfies the given initial conditions is 5. Therefore, the solution to the equation y'' - 2y' -5y =0 satisfying the conditions y(0) = 1, y'(0) = 0, and y''(0) = C is y(x) = e^x + 4e^-5x.

This result also supports the idea that the value of y'' at a specific point is determined by the values of y and y' at that point. In this case, the value of y''(0) is determined by the values of y(0) and y'(0), which are 1 and 0 respectively, and by the constant C, which is now known to be 5. This further emphasizes the importance of initial conditions in determining a unique solution to a differential equation.