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# Find the Particular Solution of the Differential Equation Tan X (Dy)By(Dx) = 2x Tan X + X^2 - Y; (Tan X Not Equal 0) Given that Y = 0 When X - CBSE (Commerce) Class 12 - Mathematics

ConceptGeneral and Particular Solutions of a Differential Equation

#### Question

Find the particular solution of the differential equation

tan x * (dy)/(dx) = 2x tan x + x^2 - y; (tan x != 0) given that y = 0 when x = pi/2

#### Solution

The given differential equation is

tan x * (dy)/(dx) = 2x tan x + x^2 - y; (tan x != 0)

=> (dy)/(dx) = 2x + x^2 cotx - y cotx

=> dy/dx + (cot x)y = 2x + x^2 cot x

This is a linear differential equation.

Here, P = cot x, Q = 2x + x2 cot x

:. I.F. = e^(int P dx) = e^(int cot s dx) = e^(log|sin x|) = sin x

The general solution of this linear differential equation is given by

y(I.F.) = ∫Q(I.F.)dx + C

=> y*sinx = int(2x + x^2 cotx) sinx dx + C

=>y*sinx = int 2xsin x dx + int x^2 cos x dx + C

y*sinx = int2x sin x dx + x^2 sinx - int 2 xsin x + C     (Applying integration by parts in the 2nd integral)

=>y*sinx = x^2 sinx +C......1

When y = 0,  x = pi/2  (Given)

:. 0 xx sin  pi/2 = pi^2/4 sin  pi/4 + C

=> C = - pi^2/4

Substituting the value of C in (1), we get

ysinx = x^2 sin x - pi^2/4

=> (x^2- y) sin x = pi^2/4

This is the particular solution of the given differential equation

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Solution Find the Particular Solution of the Differential Equation Tan X (Dy)By(Dx) = 2x Tan X + X^2 - Y; (Tan X Not Equal 0) Given that Y = 0 When X Concept: General and Particular Solutions of a Differential Equation.
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