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Calculus/succesive differenciation


Aditya Dhanraj wrote at 2015-09-02 16:56:40
After multiplying the value of Y' from 'a' we can calculate the value of (a^2)(e^ax)(sinbx).

We can also calculate the value of (ab)(e^ax)(cosbx) from the same equation.

We can replace (e^ax)(sinbx) by Y.

When we will put these values in equation of Y", our final result will be



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