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anu143

  • 2 years ago

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

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  1. him1618
    • 2 years ago
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    you need to solve the diff eqn?

  2. anu143
    • 2 years ago
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    ya by variable separable

  3. him1618
    • 2 years ago
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    x wali sari expressions ek side karlo

  4. anu143
    • 2 years ago
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    its nt workin :(

  5. him1618
    • 2 years ago
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    you have dy / (y+b) = (x+sqrt(x^2 + a^2))dx / (x^2 + a^2) isnt it?

  6. anu143
    • 2 years ago
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    correct

  7. him1618
    • 2 years ago
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    so now wheres the problem?

  8. anu143
    • 2 years ago
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    m nt gettin hw 2 go ahead

  9. him1618
    • 2 years ago
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    ok the left hand side is clear to you isnt it??

  10. anu143
    • 2 years ago
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    yup d right one s mes

  11. him1618
    • 2 years ago
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    so fr the right side break the fraction into 2 parts

  12. him1618
    • 2 years ago
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    \[\frac{x}{x^2 + a^2} + \frac{1}{\sqrt(x^2 + a^2)}\]

  13. him1618
    • 2 years ago
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    ab aaya?

  14. anu143
    • 2 years ago
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    i gt it thanks:)

  15. anu143
    • 2 years ago
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    u on fb??

  16. anu143
    • 2 years ago
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    can i hav ur id??

  17. hartnn
    • 2 years ago
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    O.o

  18. UnkleRhaukus
    • 2 years ago
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    \[(x^2+a^2)\frac{\text dy}{\text dx}=(y+b)(x+\sqrt{x^2+a^2})\] \[\frac{\text dy}{y+b}=\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\] \[\int\frac{\text dy}{y+b}=\int\frac{x+\sqrt{x^2+a^2}}{x^2+a^2}{\text dx}\]

  19. UnkleRhaukus
    • 2 years ago
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    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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