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Differential Equations

Long Answer Type

161.

For the differential equation $xy dy over dx space equals space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis straight y plus 2 right parenthesis comma$ find the solution curve passing through the point (1, -1).

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162. Find the equation of a curve passing through the point (0, – 2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

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163. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve given that it passes through ( 2, 1).

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164. Find the equation of the curve passing through the point

open parentheses 0 comma space straight pi over 4 close parentheses

whose differential equation is sin x cos y dx + cos x sin y dy = 0.

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165. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x sin x.

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166.
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Let v be volume of spherical balloon of radius r.

therefore space space space space space space space straight v space equals space 4 over 3 πr cubed

...(1)
From given condition,

dv over dt equals space straight k space space or space space straight d over dt open parentheses 4 over 3 πr cubed close parentheses space equals space straight k

open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

therefore space space fraction numerator 4 straight pi over denominator 3 end fraction. space 3 space straight r squared space dr over dt space equals space straight k space space space or space space 4 πr squared dr over dt space equals space straight k

Separating the variables and integrating, we get.

4 straight pi integral straight r squared space dr space equals space straight k space integral space dt space space space or space space space 4 space straight pi space straight r cubed over 3 space equals space straight k space straight t space plus straight c

...(2)
Now t = 0 when r = 3

therefore space space space space 4 straight pi fraction numerator left parenthesis 3 right parenthesis cubed over denominator 3 end fraction space equals space straight k cross times 0 space plus space straight c space space space rightwards double arrow space straight c space equals space 36 space straight pi

...(3)
Again t = 3 when r = 6

therefore space space fraction numerator 4 straight pi over denominator 3 end fraction left parenthesis 6 right parenthesis cubed space equals space 3 straight k space plus space 36 straight pi

open square brackets because space space of space left parenthesis 3 right parenthesis close square brackets

therefore space space space space 288 space straight pi space equals space 3 space straight k space plus space 36 space straight pi space space or space space 3 space straight k space equals space 252 space straight pi

therefore space space space straight k space equals space 84 space straight pi

Putting

straight k space equals 84 space straight pi comma space space straight c space equals space 36 space straight pi space in space left parenthesis 2 right parenthesis comma space we space get

fraction numerator 4 straight pi over denominator 3 end fraction straight r cubed equals space space 84 space straight pi space straight t space plus space 36 space straight pi space space space or space space space straight r cubed over 3 space equals space 21 space straight t space plus space 9

therefore space space space straight r cubed space equals space 63 space straight t space plus space 27 space space space space rightwards double arrow space space space space straight r space equals space left square bracket 9 space left parenthesis 7 space straight t space plus space 3 right parenthesis right square bracket to the power of 1 third end exponent

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167. In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs. 1000 double itself ?

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168. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years.

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169. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years. (e^0·5 = 1·648).

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170.

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

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Previous Year Papers

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Test Yourself

Long Answer Type

166. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

166.
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.