I have one doubt.When you get a complex number the solution will have a sine and cosine functions
like this
\(\displaystyle f(x) = c_1 e^x \cos x + c_2 e^x \sin x\)
\(\displaystyle e^x\) will take the real number and \(\displaystyle \cos x\) and \(\displaystyle \sin x\) will take the imaginary number
you have \(\displaystyle m = 1 + 3i\) and \(\displaystyle m = 1 - 3i\)
this means you have \(\displaystyle 1\) as real number and \(\displaystyle 3i\) as an imaginary number
so the solution will be
\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)
Now, you can continue with the conditions to find \(\displaystyle c_1\) and \(\displaystyle c_2\)
Also, what will be Y(0) ? I know e^x becomes 1. But what happens to cos3x and sin3x ?When you get a complex number the solution will have a sine and cosine functions
like this
\(\displaystyle f(x) = c_1 e^x \cos x + c_2 e^x \sin x\)
\(\displaystyle e^x\) will take the real number and \(\displaystyle \cos x\) and \(\displaystyle \sin x\) will take the imaginary number
you have \(\displaystyle m = 1 + 3i\) and \(\displaystyle m = 1 - 3i\)
this means you have \(\displaystyle 1\) as real number and \(\displaystyle 3i\) as an imaginary number
so the solution will be
\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)
Now, you can continue with the conditions to find \(\displaystyle c_1\) and \(\displaystyle c_2\)
When you get a complex number the solution will have a sine and cosine functions
like this
\(\displaystyle f(x) = c_1 e^x \cos x + c_2 e^x \sin x\)
\(\displaystyle e^x\) will take the real number and \(\displaystyle \cos x\) and \(\displaystyle \sin x\) will take the imaginary number
you have \(\displaystyle m = 1 + 3i\) and \(\displaystyle m = 1 - 3i\)
this means you have \(\displaystyle 1\) as real number and \(\displaystyle 3i\) as an imaginary number
so the solution will be
\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)
Now, you can continue with the conditions to find \(\displaystyle c_1\) and \(\displaystyle c_2\)
cos(3*0) = cos(0) =1 andAlso, what will be Y(0) ? I know e^x becomes 1. But what happens to cos3x and sin3x ?
You can always ignore the negative value and treat \(\displaystyle -3i\) and \(\displaystyle 3i\) as just \(\displaystyle 3i\)I have one doubt.
One of the roots is m= 1-3i which becomes sin3x. So do we ignore the negative value? What if it was 1+ 4i ? Would it been sin4x ?
Can you please tell me if my solution is correct?You can always ignore the negative value and treat \(\displaystyle -3i\) and \(\displaystyle 3i\) as just \(\displaystyle 3i\)
Why?
because \(\displaystyle 1+3i\), will give you this equation
\(\displaystyle y_1(x) = e^x \cos 3x + e^x \sin 3x \)
and
\(\displaystyle 1-3i\), will give you this equation
\(\displaystyle y_2(x) = e^x \cos -3x + e^x \sin -3x = e^x \cos 3x - e^x \sin 3x\)
the solution is their combination
\(\displaystyle u(x) = y_1(x) + y_2(x) = 2e^x \cos 3x \)
\(\displaystyle u(x) = \frac{1}{2}y_1(x) + \frac{1}{2}y_2(x) = e^x \cos 3x \)
we can have another solution as
\(\displaystyle v(x) = y_1(x) - y_2(x) = 2ie^x \sin 3x \)
\(\displaystyle v(x) = \frac{1}{2i}y_1(x) - \frac{1}{2i}y_2(x) = e^x \sin 3x \)
the general solution is just a combination of the two
\(\displaystyle y(x) = c_1 u(x) + c_2 v(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)
Your derivative is wrong
the derivative of
\(\displaystyle f(x) = c_1 e^x \cos 3x + c_2 e^x \sin 3x\)
Is Not
\(\displaystyle f'(x) = c_1 e^x(-3\sin 3x) + c_2 e^x (3\cos 3x)\)
Did you forget the product rule when you have two functions multiplying by each other such as \(\displaystyle e^x \cos 3x\)?
Yup, you got it right. Congratulations.I'm sorry. I'm a beginner. Can you please tell me if my solution is correct , or have I made further mistakes?