Advertisements
Advertisements
Question
Solve the following differential equation.
(x2 − yx2 ) dy + (y2 + xy2) dx = 0
Solution
(x2 − yx2 ) dy + (y2 + xy2) dx = 0
∴ x2 (1 - y) dy = - y2 (1 + x) dx
∴ `((1-y)/y^2)dy = - ((1+x)/x^2)dx`
Integrating on both sides, we get
`int(1/y^2- 1/y) dy = - int (1/x^2+1/x)dx`
∴ `-1/y - log |y| = - (-1/x + log | x |)+c`
∴`(-1)/y - log |y| = 1/x - log | x |+c`
∴ `log | x | - log | y | = 1/x + 1/y + c`
APPEARS IN
RELATED QUESTIONS
Integrate the function in x2 log x.
Integrate the function in x cos-1 x.
Integrate the function in ex (sinx + cosx).
`int e^x sec x (1 + tan x) dx` equals:
Integrate the following functions w.r.t. x : `e^(sin^-1x)*[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`
Evaluate the following.
`int [1/(log "x") - 1/(log "x")^2]` dx
Choose the correct alternative from the following.
`int (1 - "x")^(-2) "dx"` =
Evaluate: `int "dx"/(5 - 16"x"^2)`
`int 1/(4x + 5x^(-11)) "d"x`
Choose the correct alternative:
`intx^(2)3^(x^3) "d"x` =
`int 1/(x^2 - "a"^2) "d"x` = ______ + c
Evaluate `int 1/(4x^2 - 1) "d"x`
`int cot "x".log [log (sin "x")] "dx"` = ____________.
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.
Solve the following
`int_0^1 e^(x^2) x^3 dx`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
Evaluate:
`int e^(logcosx)dx`
Evaluate the following.
`intx^3/sqrt(1+x^4) dx`