The solution of x 3 d x d y + 4 x 2 tan y = e x sec y satisfying y ( 1 ) = 0 is sin y = e x ( x − 1 ) x − k then k =

Put sin y = t ⇒ cos y d x d y = d x d t ∴ dx dt + x 4 t = x 3 e x I.F. = e ∫ 4/ xdx = e 4 l n x = x 4 dx d ( t . x 4 ) = xe x

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MathematicsDifferential EquationLinear DE / Red. LDEMedium2 minQB

MathematicsMediuminteger

The solution of $x^{3} \frac{d y}{d x} + 4 x^{2} tan y = e^{x} sec y$ satisfying $y (1) = 0$ is $sin y = e^{x} (x - 1) x^{- k}$ then $k =$

Answer:

Solution:

Put $sin y = t \Rightarrow cos y \frac{d y}{d x} = \frac{d t}{d x}$

$∴ \frac{dt}{dx} + \frac{4}{x} t = \frac{e ^{x}}{x ^{3}}$

I.F. $= e^{\int 4/ xdx} = e^{4 l n x} = x^{4}$

$\frac{d ( t . x ^{4} )}{dx} = xe^{x}$

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Linear DE / Red. LDE

⏱ 2mℹ️ Source: QB

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