If d y d x = x y x 2 + y 2 ; y 1 = 1 ; then a value of x satisfying y x = e is:

Put y = v x d y d x = v + x d v d x v + x d v d x = v x 2 x 2 + v 2 x 2 ⇒ 1 + v 2 v 3 d v = - 1 x d x ⇒ ∫ 1 v 3 + 1 v d v = ∫ - 1 x d x ⇒ - 1 2 1 v 2 + l n v = - l n x + c ⇒ - x 2 2 y 2 = - l n y + c When x = 1 , y = 1 then - 1 2 = c ⇒ x 2 = y 2 1 + 2 l n y ⇒ x 2 = e 2 3

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Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationHomogeneous equation / Red. HDEEasy2 minPYQ_2020

MathematicsEasynumerical range

If $\frac{d y}{d x} = \frac{x y}{x^{2} + y^{2}}; y (1) = 1;$ then a value of $x$ satisfying $y (x) = e$ is:

Options:

Answer:

Solution:

Put $y = v x$

$\frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = \frac{v x^{2}}{x^{2} + v^{2} x^{2}}$

$\Rightarrow \frac{1 + v^{2}}{v^{3}} d v = - \frac{1}{x} d x$

$\Rightarrow \int (\frac{1}{v^{3}} + \frac{1}{v}) d v = \int - \frac{1}{x} d x$

$\Rightarrow \frac{- 1}{2} \frac{1}{v^{2}} + l n v = - l n x + c$

$\Rightarrow - \frac{x^{2}}{2 y^{2}} = - l n y + c$

When $x = 1, y = 1$ then

$- \frac{1}{2} = c$

$\Rightarrow x^{2} = y^{2} (1 + 2 l n y)$

$\Rightarrow x^{2} = e^{2} (3)$

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Homogeneous equation / Red. HDE

⏱ 2mℹ️ Source: PYQ_2020

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