Q. 36 I5.0( 1 Vote )

# Solve each of the following initial value problems

,y(2) = x

Answer :

,y(2) = x

It is homogeneous equation,

Put y = vx

And

So,

cosec v dv =

– log (cosce v + cot v) = – log x + c

Put y = ,x = 2,We have,

c = 0.301

now,

– log (cosec) = – log x + 0.301

Rate this question :

Find a particular solution of the differential equation (x ≠ 0), given that y = 0 when

NCERT - Mathematics Part-IIThe general solution of the differential equation e^{x} dy + (y e^{x} + 2x) dx = 0 is

Solve the differential equation

NCERT - Mathematics Part-IIFind the particular solution of the differential equation (1 + e^{2x}) dy + (1 + y^{2}) e^{x} dx = 0, given that y = 1 when x = 0.

Find the equation of the curve passing through the point whose differential equation is sin x cos y dx + cos x sin y dy = 0.

NCERT - Mathematics Part-IIShow that the general solution of the differential equation is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.

NCERT - Mathematics Part-IIFind the general solution of the differential equation

NCERT - Mathematics Part-IIFind a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

NCERT - Mathematics Part-IISolve the differential equation

NCERT - Mathematics Part-II