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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!

3 answers

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 = c_₁*sin kx + c_₂*cos kx

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