Doubt from Differential Equation

#1


Q 2

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#2

substitute \frac{dy}{dx} = z solve will get ans.. to find y you may integrate...

#3

Can you elaborate how can you find y here

#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

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#5

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#6

#7

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

#8

xdy +ydx = xdx - ydy

d(xy) = xdx - ydy

2xy = x² - y² + c

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#9

#10

Differentiate with respect to x both side.

#11

Is it option A

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#12

Yes

#13

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#14


Q 1

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#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..

#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

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#17

Nice approach, Sir.
Is a nice approach is available for the integer one too.

#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

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#19

Sir, how you concluded that m^2 = m.
According to me, we can't say about them.

#20

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