If a function y=f(x) is such that f′(x) is continuous function and satisfies (f(x))2=K+∫x0((f(t))2+(f′(t))2) dt, K ϵ R+, then
f(x) is increasing function ∀ x ϵ R
f(x) is neither even nor odd
if K=100 then f(0)=10
Differentiating the given equation 2f(x)f′(x)=(f(x))2 + (f′(x))2⇒ (f(x) – f′(x))2=0
⇒ f′(x)=f(x)⇒f(x)=cex
∵ f(0)=√K⇒ c=√K⇒f(x)=√Kex