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

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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.

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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?

Let y denote the bacteria at any time t.
From the given condition,

dy over dt space equals space straight k space straight y comma space

where k is the constant of proportionality.
Separting the variables and integrating, we get,

integral 1 over straight y dy space equals space straight k integral dt

therefore space space space space space space space log space straight y space equals space straight k space straight t space space plus space straight c

...(1)
Let

straight y subscript 0

be bacteria at time t = 0

therefore space space space space space space space space space space space log space straight y subscript 0 space equals space 0 plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space log space straight y subscript 0 therefore space space space space from space left parenthesis 1 right parenthesis comma space log space straight y space equals space straight k space straight t space plus space log space straight y subscript 0 space space space space space space space space space space space space space space rightwards double arrow space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space straight k space straight t space space space space space... left parenthesis 2 right parenthesis

Now,

straight y space equals space straight y subscript 0 plus 10 over 100 straight y subscript 0 space equals space 11 over 10 straight y subscript 0 space space space when space straight t space equals space 2

therefore space space space from space left parenthesis 2 right parenthesis comma space space log open parentheses fraction numerator 11 space straight y subscript 0 over denominator 10 space straight y subscript 0 end fraction close parentheses space equals space 2 space straight k space space rightwards double arrow space space space straight k space equals space 1 half log space open parentheses 11 over 10 close parentheses

therefore space space from space left parenthesis 2 right parenthesis comma space space log space open parentheses straight y over straight y subscript 0 close parentheses space equals space 1 half log space open parentheses 11 over 10 close parentheses. space straight t subscript 1

...(3)
Let the bacteria be 2,00,000 from 1,00,000 in

straight t subscript 1

hours
i.e.

straight y space equals space 2 space straight y subscript 0 space space space space when space straight t space space equals space straight t subscript 1

therefore space space space space from space left parenthesis 3 right parenthesis comma space space log space open parentheses fraction numerator 2 space straight y subscript 0 over denominator straight y subscript 0 end fraction close parentheses space equals space 1 half log space open parentheses 11 over 10 close parentheses. space straight t subscript 1

therefore space space space space space space log space 2 space equals space 1 half log open parentheses 11 over 10 close parentheses space straight t subscript 1 therefore space space space space space space space space space straight t subscript 1 space equals space fraction numerator 2 space log space 2 over denominator log space open parentheses begin display style 11 over 10 end style close parentheses end fraction space hours. space

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