JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 8)
Let y = y(x) be the solution of the differential
equation (x $$-$$ x3)dy = (y + yx2 $$-$$ 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :
equation (x $$-$$ x3)dy = (y + yx2 $$-$$ 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :
4
12
8
16
Explanation
$$(x - {x^3})dy = (y + y{x^2} - 3{x^4})dx$$
$$ \Rightarrow xdy - ydx = (y{x^2} - 3{x^4})dx + {x^3}dy$$
$$ \Rightarrow {{xdy - ydx} \over {{x^2}}} = (ydx + xdy) - 3{x^2}dx$$
$$ \Rightarrow d\left( {{y \over x}} \right) = d(xy) - d({x^3})$$
Integrate
$$ \Rightarrow {y \over x} = xy - {x^3} + c$$
given f(3) = 3
$$ \Rightarrow {3 \over 3} = 3 \times 3 - {3^3} + c$$
$$ \Rightarrow c = 19$$
$$\therefore$$ $${y \over x} = xy - {x^3} + 19$$
at $$x = 4,{y \over 4} = 4y - 64 + 19$$
$$15y = 4 \times 45$$
$$ \Rightarrow y = 12$$
$$ \Rightarrow xdy - ydx = (y{x^2} - 3{x^4})dx + {x^3}dy$$
$$ \Rightarrow {{xdy - ydx} \over {{x^2}}} = (ydx + xdy) - 3{x^2}dx$$
$$ \Rightarrow d\left( {{y \over x}} \right) = d(xy) - d({x^3})$$
Integrate
$$ \Rightarrow {y \over x} = xy - {x^3} + c$$
given f(3) = 3
$$ \Rightarrow {3 \over 3} = 3 \times 3 - {3^3} + c$$
$$ \Rightarrow c = 19$$
$$\therefore$$ $${y \over x} = xy - {x^3} + 19$$
at $$x = 4,{y \over 4} = 4y - 64 + 19$$
$$15y = 4 \times 45$$
$$ \Rightarrow y = 12$$
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