Differential Equation: y' = 2(3x + y)^2 - 1
How can i solve this problem y' = 2(3x+y)2-1 when x = 0, y =0. Can someone show me the solution ?
July 2016
More Topics on Dynamics
Please upload more examples and answers on Dynamics sirs maam Thanks in advance
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DE exact equations: (3 + y + 2y^2 sin^2 x) dx + (x + 2xy - y sin 2x) dy = 0
(3 + y + 2y^2 sin ^ 2 ( x) ) dx + ( x + 2xy - y sin 2x) dy = 0
DE Order one: (xy^2 + x - 2y + 3) dx + x^2 ydy = 2(x + y) dy
(xy^2 + x - 2y + 3) dx + x^2 ydy = 2(x+y) dy
equations of order one: (1 - xy)^2 dx + [ y^2 + x^2 (1 - xy)^(-2) ] dy = 0
(1 - xy ) ^ 2 (dx) + [ y^2 + x^2 ( 1 - xy ) ^ -2 ] dy =0
DE: 2xy dx + (y^2 - x^2) dy = 0
2xydx + ( y^2 - x^2 ) dy = 0
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DE: 2xy dx + (y^2 + x^2) dy = 0
2xydx + ( y^2 + x^2 ) dy = 0
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exact DE: [ 2x + y cos (x^2) - 2xy + 1 ] dx + [ sin (x^2) - x^2 ] dy = 0
[ 2x + y cos (x^2) - 2xy + 1 ] dx + [ sin (x^2) - x^2 ] dy = 0
Differential Equation: [ e^(2y) - y cos (xy) ] dx - y(1 - x^2) dy = 0
Please help me again to solve this.
(e2y-ycosxy)dx-y(1-x2)dy=0
Differential Equations
How can we simplify ln(9m^2 n -mn^3) +x +y = c ?
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