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The solution of the differential equation $$x{{dy} \over {dx}} + 2y$$ = x2 (x $$ \ne $$ 0) with y(1) = 1, is :

JEE · Math · previous-year question

  1. A.$$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$$
  2. B.$$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$$
  3. C.$$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$correct
  4. D.$$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$$

Answer

C. $$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$

Explanation

$$x{{dy} \over {dx}} + 2y$$ = x2 $$ \Rightarrow $$ $${{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x$$ $$ \therefore $$ I.F = $${e^{\int {{2 \over x}dx} }}$$ = x2 $$ \therefore $$ The solution is yx2 = $$\int {{x^3}dx} $$ $$ \Rightarrow $$ yx2 = $${{{x^4}} \over 4} + C$$ .....(1) As y(1) = 1 $$ \therefore $$ when x = 1 then y = 1. Putting the value of x and y in equation (1), we get 1 = $${1 \over 4} + C$$ $$ \Rightarrow $$ C = $${3 \over 4}$$ $$ \therefore $$ Required solution yx2 = $${{{x^4}} \over 4} + {3 \over 4}$$ $$ \Rightarrow $$ $$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$

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