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Question
For the differential equation find a particular solution satisfying the given condition:
`dy/dx - y/x + cosec (y/x) = 0; y = 0` when x = 1
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Solution
Given differential equation
`dy/dx - y/x + cosec (y/x)` = 0 ....(i)
`dy/dx = y/x - cosec (y/x)`
Clearly, this equation is a differential equation.
∴ Putting y = vx
`v + x (dv)/dx` = v - cosec v in equation (i)
`=> x (dv)/dx` = - cosec v
sin v dv = `- 1/x` dx
On integrating,
`- int sin v dv = int 1/x dx`
cos v = log x + C
So on putting `(y/x)` in place of v,
`cos (y/x) = log |x| + C` ....(ii)
Given y = 0 if x = 1 ...[from equation (ii)]
cos 0 = log 1 + C
⇒ C = 1
Putting this value of C in equation (ii),
`cos (y/x) = log |x| + log |e|`
`cos (y/x) = log |ex|`
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An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form `dy/dx` = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λn F(x, y). To solve a homogeneous differential equation of the type `dy/dx` = F(x, y) = `g(y/x)`, we make the substitution y = vx and then separate the variables. |
Based on the above, answer the following questions:
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