6 - Applications of integrations Questions Answers

this is a special type of differential equation in which p = dy/dx

its solution will be y = cx+(a/c)  here c is a constant

on applying p = dy/dx

(ydx-xdy)/dx = (xdx+ydy)/dx

or ydx-xdy = xdx+ydy

or dy/dx = (y-x)/(y+x)

now it is homogeneous

according to the given condition f(nx) = n f(x)

₀òⁿ f(x) dx

consider x = ny so this integration will become ₀ò¹ f(ny) ndy = n² ₀ò¹ f(y) dy = n² ₀ò¹ f(x) dx

now by using these conditions I₁ = 1³ ₀ò¹ f(x) dx similarly others 

put these values and get the answer

 let I = ₀ò^p q ln sin q dq

on aplying property of definite integral

I =  ₀ò^p (π-q) ln sin q dq

so 2 I =  ₀ò^p π ln sin q dq 

or I = π/2  ₀ò^p  ln sin q dq

or I = 2π/2  ₀ò^p/2 ln sin q dq    this is due to property

similarly solve  ₀ò^p q² ln sin q dq  and   then ₀ò^p q³ ln sin q dq 

finally solve the right hand side of the equation to prove LHS = RHS

f(x) = 1/2 ₀ò^x (x-t)² g(t) dt

or f(x) = 1/2 ₀ò^x x² g(t) dt + 1/2 ₀ò^x t² g(t) dt - ₀ò^x x t g(t) dt 

or f(x) = 1/2 x² ₀ò^x  g(t) dt + 1/2 ₀ò^x t² g(t) dt - x ₀ò^x  t g(t) dt

now use newton leibniz formula for middle function and product rule for first and last functions for finding f^' (x)

similarly get f^'' (x)