The Principle of Superposition for Homogeneous Equations (DiffEq)

[_N3WTON_]

Homework Statement

Verify that e^x and e^-x and any linear combination c_1e^x + c_2e^{-x} are all solutions of the differential equation:
y'' - y = 0
Show that the hyperbolic sine and cosine functions, sinhx and coshx are also solutions

Homework Equations

Principle of Superposition for Homogeneous Equations
y'' + p(x)y' + q(x)y = 0
y(x) = c_1y_1(x) = c_2y_2(x)

The Attempt at a Solution

I am not having any trouble on the first part, here is my solution:
y_1(x) = e^x
y_2(x) = e^{-x}
y'' - y = 0
y = c_1e^x + c_2e^{-x}
y' = c_1e^x - c_2e^{-x}
y'' = c_1e^x + c_2e^{-x}
(c_1e^x + c_2e^{-x}) - (c_1e^x + c_2e^{-x}) = 0
0 = 0
Now, on the second part of the problem I run into problems, here is what I have so far:
y_1(x) = sinh(x)
y_2(x) = cosh(x)
y = c_1cosh(x) + c_2sinh(x)
y' = -c_1sinh(x) + c_2cosh(x)
y'' = -c_1cosh(x) - c_2sinh(x)
(-c_1cosh(x) - c_2sinh(x)) - (c_1cosh(x) + c_2sinh(x)) = 0
However, the last equation is not true and I am not sure where I went wrong...

 

[Mark44]

Homework Statement

Verify that e^x and e^-x and any linear combination c_1e^x + c_2e^{-x} are all solutions of the differential equation:
y'' - y = 0
Show that the hyperbolic sine and cosine functions, sinhx and coshx are also solutions

Homework Equations

Principle of Superposition for Homogeneous Equations
y'' + p(x)y' + q(x)y = 0
y(x) = c_1y_1(x) = c_2y_2(x)

The Attempt at a Solution

I am not having any trouble on the first part, here is my solution:
y_1(x) = e^x
y_2(x) = e^{-x}
y'' - y = 0
y = c_1e^x + c_2e^{-x}
y' = c_1e^x - c_2e^{-x}
y'' = c_1e^x + c_2e^{-x}
(c_1e^x + c_2e^{-x}) - (c_1e^x + c_2e^{-x}) = 0
0 = 0
Now, on the second part of the problem I run into problems, here is what I have so far:
y_1(x) = sinh(x)
y_2(x) = cosh(x)
y = c_1cosh(x) + c_2sinh(x)
y' = -c_1sinh(x) + c_2cosh(x)
y'' = -c_1cosh(x) - c_2sinh(x)
(-c_1cosh(x) - c_2sinh(x)) - (c_1cosh(x) + c_2sinh(x)) = 0
However, the last equation is not true and I am not sure where I went wrong...

Your derivative for cosh(x) is wrong. d/dx(cosh(x)) = sinh(x).

 

[LCKurtz]

The derivatives of hyperbolic sinh cosh functions don't have a minus sign like the ordinary sines and cosines do.

 

[_N3WTON_]

thank you both...I was so confused for a moment XD