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Question Number 208805 by Mastermind last updated on 23/Jun/24

Integrate:  (xdz − zdx) − a^2 (2xzdz − z^2 dx) + 2x^3  = 0

$$\mathrm{Integrate}: \\ $$$$\left(\mathrm{xdz}\:−\:\mathrm{zdx}\right)\:−\:\mathrm{a}^{\mathrm{2}} \left(\mathrm{2xzdz}\:−\:\mathrm{z}^{\mathrm{2}} \mathrm{dx}\right)\:+\:\mathrm{2x}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$

Commented by mr W last updated on 23/Jun/24

things like f(x)dx+g(x)=0 make no  sense. you have such a thing. please  check the question again.

$${things}\:{like}\:{f}\left({x}\right){dx}+{g}\left({x}\right)=\mathrm{0}\:{make}\:{no} \\ $$$${sense}.\:{you}\:{have}\:{such}\:{a}\:{thing}.\:{please} \\ $$$${check}\:{the}\:{question}\:{again}. \\ $$

Commented by Mastermind last updated on 23/Jun/24

Ok thank you but there′s original ques.    Here′s the question  Solve:  px(z − 2y^2 ) = (z − qy)(z − y^2  − 2x^3 )    It is a P.D.E question. Thank you

$$\mathrm{Ok}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{original}\:\mathrm{ques}. \\ $$$$ \\ $$$$\mathrm{Here}'\mathrm{s}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{Solve}: \\ $$$$\mathrm{px}\left(\mathrm{z}\:−\:\mathrm{2y}^{\mathrm{2}} \right)\:=\:\left(\mathrm{z}\:−\:\mathrm{qy}\right)\left(\mathrm{z}\:−\:\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{2x}^{\mathrm{3}} \right) \\ $$$$ \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{P}.\mathrm{D}.\mathrm{E}\:\mathrm{question}.\:\mathrm{Thank}\:\mathrm{you} \\ $$

Commented by Mastermind last updated on 23/Jun/24

It is P.D.E, under method of multiplier (Hint)

$$\mathrm{It}\:\mathrm{is}\:\mathrm{P}.\mathrm{D}.\mathrm{E},\:\mathrm{under}\:\mathrm{method}\:\mathrm{of}\:\mathrm{multiplier}\:\left(\mathrm{Hint}\right) \\ $$

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