Question

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Answer the question which is in attachment.

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Answer 1

Answer:

Find general and singular solution of px=3p2ey+1, p=dydx

My attempt:-

px=(3p2ey+1),x=3pey+1p

differentiating w.r.to y

1/p=3eydpdy+3pey−1p2dpdy ⟹1−3p2eyp=3eyp2−1p2dpdy ⟹dpp+dy=0,p=ce−y,

c is arbitrary constant

putting this value of p in given equation, i got

3p2ey−px+1=0⟹3c2e−y−ce−yx+1=0

is required general solution

But textbook exercise answer is 3c2e3y−xcey+1=0 which means p=cey

I got p=ce−y. i verified but could not find mistake. Pls have a look at this.

Answer 2

Answer:

bro

please answer for my questions in physics