JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 19)
The solution of the differential equation
$$x{{dy} \over {dx}} + 2y$$ = x2 (x $$ \ne $$ 0) with y(1) = 1, is :
$$x{{dy} \over {dx}} + 2y$$ = x2 (x $$ \ne $$ 0) with y(1) = 1, is :
$$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$$
$$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$$
$$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$
$$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$$
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 = $${{{x^4}} \over 4} + {3 \over 4}$$
$$ \Rightarrow $$ $$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$
$$ \Rightarrow $$ $${{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x$$
$$ \therefore $$ 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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