i just worked out that a=1 b=4 and c=0
is that right ..did anyone else get that?
i want to find the values of the real constants a,b,c for which x(t) = ae^(bt) +ct is a solution to the ODE d^2/dt^2 = 16e^(4t) / xThis is what i have done so far
i differentiated x to get abe^(bt) + c
then i differentialiated that again to get (ab^2)*e^(bt)
then i substituted x and x" into the ODE given to get
(ab^2)*e^(bt) = 16e^(4t) / ae^(bt) +ct
this is where i got stuck..am i on the right path..from this part i need to find a b and c..any help thankz
Now we are told that,
Is some solution.
Then by substituting it into the differencial equation we have equality.
First we find the first derivative,
Note the left hand side has some multiple of while the right does not. Thus to match we need it to be zero. Thus, one of the constants (at least) is zero, that is .
Thus it dies,
We see that it cannot be because they must be non-zero in this equation. Thus, .
The exponents need to match, for equality, that is,
Then by substituting for this value we have,
We immediately see that .
Thus, there are two possibilities,
For the right and left hand sides to be equal you make the coefficients equal to each other.
1)There is a on left side and there is on right side. To be equal the coefficient must match, that is, .
2)There is on left side and there is on right side. To be equal the coefficient must match, that is, . Thus, .
3)And there is not on left hand side. Thus,