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

Mathematics
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you need to solve the diff eqn?
ya by variable separable
x wali sari expressions ek side karlo

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Other answers:

its nt workin :(
you have dy / (y+b) = (x+sqrt(x^2 + a^2))dx / (x^2 + a^2) isnt it?
correct
so now wheres the problem?
m nt gettin hw 2 go ahead
ok the left hand side is clear to you isnt it??
yup d right one s mes
so fr the right side break the fraction into 2 parts
\[\frac{x}{x^2 + a^2} + \frac{1}{\sqrt(x^2 + a^2)}\]
ab aaya?
i gt it thanks:)
u on fb??
can i hav ur id??
O.o
\[(x^2+a^2)\frac{\text dy}{\text dx}=(y+b)(x+\sqrt{x^2+a^2})\] \[\frac{\text dy}{y+b}=\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\] \[\int\frac{\text dy}{y+b}=\int\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\]
then uses these standard integrals for \(\alpha,\beta,\gamma\in\mathbb R\) \[\int\frac{\text du}{u+\alpha}=\ln|u+\alpha|\] \[\int\frac{v}{v^2+\beta^2}\text dv=\frac12\ln|v^2+\beta^2|\] \[\int\frac1{\sqrt{w^2\pm \gamma^2}}\text dw=2\sqrt{ x\pm\gamma}=\]

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