# Doubt from Differential Equation

**Amay_2018**#4

The trick is to consider

ydy/dx = z

Let us apply product rule on Z

dz/dx = y d²y/(dx²) + (dy/dx)² =0 according to problem

Therefore

Z = C (since dz/dx =0)

Hence

ydy/dx = C

Putting x = 0

C =8

Z = 8 = ydy/dx

therefore y²/2 = 8x + D

y² = 16x + D

**Sneha_2021**#7

exam technique go by option check b will be ans.. other method forming differential no pen paper so cant send solution.. sorry.

**Sneha_2021**#15

sri dharacharya formula and after that solve diffrencial equation.. one another concept approch is also given in airihant i dont remember name of that concept you may see it in airihant or ask @Azimuddin_2019 sir i read in some post of him..

**Hari_Shankar**#16

#1: We have \displaystyle x \int_0^x (f(t))^2 \ \mathrm{dt} = \left(\int_0^x f(t) \ \mathrm{dt} \right)^2

or \displaystyle \int_0^x \mathrm{dt} \int_0^x (f(t))^2 \ \mathrm{dt} = \left(\int_0^x f(t) \ \mathrm{dt} \right)^2.

By Cauchy-Schwarz we have \displaystyle \int_0^x \mathrm{dt} \int_0^x (f(t))^2 \ \mathrm{dt} \ge\left(\int_0^x f(t) \ \mathrm{dt} \right)^2.

Since equality occurs we must have \displaystyle \int_0^x \mathrm{dt} = \lambda \int_0^x f(t) \ \mathrm{dt} or f(x)=\dfrac{1}{\lambda}

But f(0)=0. Hence f(x) \equiv 0

**Hari_Shankar**#18

If y=mx+c then the given equation is m^2x+m=mx+c. Hence m^2=m i.e. m=0,1

So two lines