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प्रश्न
Solve: `y + "d"/("d"x) (xy) = x(sinx + logx)`
उत्तर
The given differential equation is `y + "d"/("d"x) (xy) = x(sinx + logx)`
⇒ `y + x * ("d"y)/("d"x) + y = x(sinx + logx)`
⇒ `x ("d"y)/("d"x) = x(sinx + logx) - 2y`
⇒ `("d"y)/("d"x) = (sinx + logx) - (2y)/x`
⇒ `("d"y)/("d"x) + 2x y = (sinx + logx)`
Here, P = `2/x` and Q = `(sinx + log x)`
Integrating factor I.F. = `"e"^(intPdx)`
= `"e"^(int 2/x dx)`
= `"e"^(2logx)`
= `"e"^(log x^2)`
= x2
∴ Solution is `y xx "I"."F". = int "Q"."I"."F". "d"x + "c"`
⇒ `y . x^2 = int (sinx + logx)x^2 "d"x + "c"` ....(1)
Let I = `int (sinx + logx)x^2 "d"x`
= `int_"I"x^2 sinx "d"x + int_"iII"^(x^2) log x "d"x`
= `[x^2 . int sinx "d"x - int("D"(x^2) . int sinx "d"x)"d"x] + [logx . intsinx "d"x - int ("D"(logx) . intx^2 "d"x)"d"x]`
= `[x^2(-cosx) -2 int - x cosx "d"x] + [logx . x^3/3 - int 1/x * x^3/3 "d"x]`
= `[-x^2 cosx + 2(xsinx - int1 .sinx "d"x)] + [x^3/3 log x - 1/3 int x^2 "d"x]`
= `-x^2cosx + 2x sinx + 2cosx + x^3/3 log x - 1/9 x^3`
Now from equation (1) we get,
`y . x^2 = -x^2 cosx + 2x sinx + 2cosx + x^3/3 log x - 1/9 x^3 + "c"`
∴ y = `-cosx + (2sinx)/x + (2cosx)/x^2 + (xlogx)/3 - 1/9 x + "c" .x^-2`
Hence, the required solution is `-cosx + (2sinx)/x + (2cosx)/x^2 + (xlogx)/3 - 1/9 x + "c" .x^-2`
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