Solve)- (2x+y+1)dx+(4x+2y-1)dy=0
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Let z = 2x + y.
So, dz = 2 dx + dy.
Hence, (2x+y + 1) dx + (4x+2y - 1) dy = 0
==> ((2x+y) + 1) dx + (2(2x+y) - 1) dy = 0
==> (z + 1) dx + (2z - 1) (dz - 2 dx) = 0
==> (-3z + 3) dx + (2z - 1) dz = 0, which is now easy to solve by separation of variables.
3 dx = (2z - 1) dz/(z - 1)
.......= [2 + 1/(z-1)] dz.
Integrate both sides:
3x = 2z + ln|z-1| + C
==> 3x = 2(2x + y) + ln|2x + y - 1| + C.
So, dz = 2 dx + dy.
Hence, (2x+y + 1) dx + (4x+2y - 1) dy = 0
==> ((2x+y) + 1) dx + (2(2x+y) - 1) dy = 0
==> (z + 1) dx + (2z - 1) (dz - 2 dx) = 0
==> (-3z + 3) dx + (2z - 1) dz = 0, which is now easy to solve by separation of variables.
3 dx = (2z - 1) dz/(z - 1)
.......= [2 + 1/(z-1)] dz.
Integrate both sides:
3x = 2z + ln|z-1| + C
==> 3x = 2(2x + y) + ln|2x + y - 1| + C.
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