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Question-109400




Question Number 109400 by mohammad17 last updated on 23/Aug/20
Commented by mohammad17 last updated on 23/Aug/20
help me sir?
$${help}\:{me}\:{sir}? \\ $$
Commented by bobhans last updated on 23/Aug/20
using Exact diff equation
$${using}\:{Exact}\:{diff}\:{equation} \\ $$
Commented by mohammad17 last updated on 23/Aug/20
sir the equation is not exact ?  are you can help me ?
$${sir}\:{the}\:{equation}\:{is}\:{not}\:{exact}\:? \\ $$$${are}\:{you}\:{can}\:{help}\:{me}\:? \\ $$
Commented by mohammad17 last updated on 23/Aug/20
no one here can you solve this ?
$${no}\:{one}\:{here}\:{can}\:{you}\:{solve}\:{this}\:? \\ $$
Answered by ajfour last updated on 24/Aug/20
(1+x^2 y)dx+(x^2 +1)ydy=0  (dy/dx)=−(1/y)(((1+x^2 y)/(1+x^2 )))  .......
$$\left(\mathrm{1}+{x}^{\mathrm{2}} {y}\right){dx}+\left({x}^{\mathrm{2}} +\mathrm{1}\right){ydy}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\mathrm{1}}{{y}}\left(\frac{\mathrm{1}+{x}^{\mathrm{2}} {y}}{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$$……. \\ $$

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