Asked by Ashley
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!
I tried integrating the differential equation, but only resulted with squares and cubes of y and x.
Please guide!
Answers
Answered by
oobleck
It is clearly NOT a solution
y = 1
y" = 0
y"+4y = 4, not 0
y = 1
y" = 0
y"+4y = 4, not 0
Answered by
Ashley
Thanks a lot!
Answered by
oobleck
Just for reference, you should not have been getting powers of x and y.
y" + k^2y = 0
has the general solution
y = c<sub><sub>1</sub></sub>*sin kx + c<sub><sub>2</sub></sub>*cos kx
y" + k^2y = 0
has the general solution
y = c<sub><sub>1</sub></sub>*sin kx + c<sub><sub>2</sub></sub>*cos kx
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