The solution of the differential equation d x d y = x x + y satisfying the condition y ( 1 ) = 1 is

y = vx d x d y = v + x d x d v v + x d x d v = 1 + v ⇒ d v = x d x ∴ v = lo g x + c ⇒ x y = lo g x + c Since, y ( 1 ) = 1 , we have y = x lo g x + x

Home
PYQ PracticeFree
JEE Rank Predictor
Answer Key

Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationHomogeneous equation / Red. HDEMedium2 minPYQ_2008

MathematicsMediumnumerical range

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

Options:

Answer:

Solution:

$y = vx$ $\frac{d y}{d x} = v + x \frac{d v}{d x} v + x \frac{d v}{d x} = 1 + v \Rightarrow d v = \frac{d x}{x} ∴ v = lo g x + c \Rightarrow \frac{y}{x} = lo g x + c$ Since, $y (1) = 1$ , we have $y = x lo g x + x$

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

⏱ 2mℹ️ Source: PYQ_2008

Doubts & Discussion

Loading discussions...