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x(x^(2) - 1)(dy)/(dx) = 1, y = 0 when x ...

`x(x^(2) - 1)(dy)/(dx) = 1, y = 0` when x = 2.

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The correct Answer is:
`y = 1/(2) log (x^(2) - 1)/(x^(2)) -(1)/(2) log(3)/(4)`
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NCERT BANGLISH-DIFFERENTIAL EQUATIONS-EXERCISE - 9.4
  1. sec^(2)x tan y dx + sec^(2) y tan x dy = 0

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  2. (e^(x) + e^(-x))dy - (e^(x) - e^(-x)) dx = 0

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  3. (dy)/(dx) = (1 + x^(2))(1 + y^(2))

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  4. y log y dx - x dy = 0

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  5. x^(5)(dy)/(dx) = - y^(5)

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  6. For each of the differential equation find the general solution (dy)/(...

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  7. e^(x) tan y dx + (1 - e^(x))sec^(2)y dy = 0

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  8. For each of the differential equations in Exercises 11 to 14, find a p...

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  9. x(x^(2) - 1)(dy)/(dx) = 1, y = 0 when x = 2.

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  10. cos ((dy)/(dx)) = a (a ne R), y = 1 when x = 0

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  11. (dy)/(dx) = y tan x, y = 1 when x = 0

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  12. Find the equation of a curve passing through the point (0,0) and whose...

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  13. For the differential equation xy(dy)/(dx) = (x + 2)( y + 2), find the ...

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  14. Find the equation of a curve passing through the point (0, -2) given t...

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  15. At any point (x,y) of a curve, the slope of the tangent is twice the s...

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  16. The volume of spherical balloon being inflated changes at a constant r...

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  17. In a bank, principal increases continuously at the rate of r% per year...

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  18. In a bank, principal increases continuously at the rate of 5% per year...

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  19. In a culture, the bacteria count is 1,00,000. The number is increased...

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  20. The general solution of the differential equation (dy)/(dx) = e^(x + y...

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