Exponentially sinusoidal

Calculus Level 5

Consider the function

f ( y ) = 0 [ e ( x 2 ) sin ( 2 x y ) ] d x f(y) = \displaystyle \int_{0} ^{\infty} [e^{(-x^2)} \sin(2xy)] \, dx

Let A A , B B , C C and D D respectively be the coefficients of d ( f ( y ) ) d y \frac{d(f(y))}{dy} , y f ( y ) yf(y) , y y and y 0 y^0 in the first order differential equation of f ( y ) f(y) .

A d ( f ( y ) ) d y + B y f ( y ) + C y + D = 0 A\frac{d(f(y))}{dy}+ Byf(y)+Cy+D=0 where it is also given that A A is positive .

further f ( y ) f(y) can be expressed as follows

f ( y ) = 0 y [ e ( p x 2 ) ( q y 2 ) ] d x f(y)= \int_{0} ^{y} [e^{(px^2)-(qy^2)}] \, dx

Then evaluate A + B + C + D + p + q A+B+C+D+p+q

Note that A , B , C , D , p , q A, B, C, D, p, q are real numbers and they have no common factor


The answer is 4.

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1 solution

for Leibnitz Rule see below :-

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