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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you need to solve the diff eqn?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya by variable separable

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x wali sari expressions ek side karlo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you have dy / (y+b) = (x+sqrt(x^2 + a^2))dx / (x^2 + a^2) isnt it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so now wheres the problem?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0m nt gettin hw 2 go ahead

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok the left hand side is clear to you isnt it??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup d right one s mes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so fr the right side break the fraction into 2 parts

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x}{x^2 + a^2} + \frac{1}{\sqrt(x^2 + a^2)}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[(x^2+a^2)\frac{\text dy}{\text dx}=(y+b)(x+\sqrt{x^2+a^2})\] \[\frac{\text dy}{y+b}=\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\] \[\int\frac{\text dy}{y+b}=\int\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0then uses these standard integrals for \(\alpha,\beta,\gamma\in\mathbb R\) \[\int\frac{\text du}{u+\alpha}=\lnu+\alpha\] \[\int\frac{v}{v^2+\beta^2}\text dv=\frac12\lnv^2+\beta^2\] \[\int\frac1{\sqrt{w^2\pm \gamma^2}}\text dw=2\sqrt{ x\pm\gamma}=\]
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