How do we show that the function y(x)=1 is a solution of y'' + 4y = 0 ?

I tried integrating the differential equation, but only resulted with squares and cubes of y and x.

Please guide!

It is clearly NOT a solution

y = 1
y" = 0
y"+4y = 4, not 0

Thanks a lot!

Just for reference, you should not have been getting powers of x and y.

y" + k^2y = 0
has the general solution
y = c1*sin kx + c2*cos kx

To determine if the function y(x) = 1 is a solution of the differential equation y'' + 4y = 0, you need to substitute it into the equation and see if it satisfies the equation.

First, find the second derivative of y(x) = 1. Since y is a constant, its derivative is zero, and the second derivative is also zero.

Now, substitute y = 1 and its derivatives into the differential equation:

y'' + 4y = 0
0 + 4(1) = 0
4 = 0

Since 4 is not equal to 0, the equation is not satisfied.

Therefore, the function y(x) = 1 is not a solution to the differential equation y'' + 4y = 0.

When integrating the differential equation, it is important to handle the integration correctly. In this case, you do not need to integrate the equation, as you are verifying if a specific function is a solution.

To find the general solutions to the differential equation y'' + 4y = 0, you would need to integrate twice. However, if you are trying to verify a specific function as a solution, simply substitute it into the equation and check if it satisfies the equation.