Question

wts d integration of (x^2+a^2)dy/dx=(y+b)(x+sqrt(x^2+a^2))

Answer 1

you need to solve the diff eqn?

Answer 2

ya by variable separable

Answer 3

x wali sari expressions ek side karlo

Answer 4

its nt workin :(

Answer 5

you have

dy / (y+b) = (x+sqrt(x^2 + a^2))dx / (x^2 + a^2)
isnt it?

Answer 6

correct

Answer 7

so now wheres the problem?

Answer 8

m nt gettin hw 2 go ahead

Answer 9

ok the left hand side is clear to you isnt it??

Answer 10

yup d right one s mes

Answer 11

so fr the right side

break the fraction into 2 parts

Answer 12

\[\frac{x}{x^2 + a^2} + \frac{1}{\sqrt(x^2 + a^2)}\]

Answer 13

ab aaya?

Answer 14

i gt it thanks:)

Answer 15

u on fb??

Answer 16

can i hav ur id??

Answer 17

O.o

Answer 18

\[(x^2+a^2)\frac{\text dy}{\text dx}=(y+b)(x+\sqrt{x^2+a^2})\]
\[\frac{\text dy}{y+b}=\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\]
\[\int\frac{\text dy}{y+b}=\int\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\]

Answer 19

then uses these standard integrals
for \(\alpha,\beta,\gamma\in\mathbb R\)

\[\int\frac{\text du}{u+\alpha}=\ln|u+\alpha|\]
\[\int\frac{v}{v^2+\beta^2}\text dv=\frac12\ln|v^2+\beta^2|\]
\[\int\frac1{\sqrt{w^2\pm \gamma^2}}\text dw=2\sqrt{ x\pm\gamma}=\]