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प्रश्न
Find the general solution of the differential equation:
`(x^2 + 1) dy/dx + 2xy = sqrt(x^2 + 4)`
उत्तर
Given equation is
`(x^2 + 1) dy/dx + 2xy = sqrt(x^2 + 4)`
`\implies dy/dx + (2x)/((x^2 + 1)) xx y = sqrt(x^2 + 4)/((x^2 + 1))`
It is of the form,
`dy/dx + Py` = Q
∴ I.F. = `e^(int P.dx)`
= `e^(int((2x)/(x^2 + 1))dx)`
Let x2 + 1 = t
2x dx = dt
∴ I.F. = `e^(int dt/t)`
= elog t
= t
= x2 + 1
General solution is given as
(I.F.)y = `int (I.F.) Q dx + C`
`\implies` (x2 + 1)y = `int (x^2 + 1) sqrt(x^2 + 4)/((x^2 + 1)) dx + C`
`\implies` (x2 + 1)y = `int sqrt(x^2 + 4) dx + C`
`\implies` (x2 + 1)y = `int sqrt((x)^2 + 2^2) dx + C`
`\implies` (x2 + 1)y = `x/2 (sqrt(x^2 + 4) + 4/2 log |x + sqrt(x^2 + 4)|) + C`
`\implies` (x2 + 1)y = `x/2 sqrt(x^2 + 4) + 2 log |x + sqrt(x^2 + 4) | + C`
is the required solution of given differential equation.
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