Differential Equations - Exact Equations
tutorial.math.lamar.edu › Classes › DEOct 08, 2018 · d d x ( Ψ ( x, y ( x))) = 0 d d x ( Ψ ( x, y ( x))) = 0. Now, if the ordinary (not partial…) derivative of something is zero, that something must have been a constant to start with. In other words, we’ve got to have Ψ ( x, y) = c Ψ ( x, y) = c. Or, y 2 + ( x 2 + 1) y − 3 x 3 = c y 2 + ( x 2 + 1) y − 3 x 3 = c.
EXAMPLE: EXACT DIFFERENTIAL EQUATIONS
math.jhu.edu › f18 › ExampleProblemsEXAMPLE: EXACT DIFFERENTIAL EQUATIONS 110.302 DIFFERENTIAL EQUATIONS PROFESSOR RICHARD BROWN Problem. Solve the Initial Value Problem 2x+ y2 + 2xy dy dx = 0, y(1) = 1. Strategy. Solving this ODE with an initial point means nding the particular solution to the ODE that passes through the point (1;1) in the ty-plane. Here we show that the ODE is
Examples On Exact Differential Equations | What is Examples ...
www.cuemath.com › jee › examples-on-exactExamples On Exact Differential Equations. Example – 16. Solve the DE 2xydx +(x2 +3y2)dy = 0. 2 x y d x + ( x 2 + 3 y 2) d y = 0. Solution: First of all, notice that this DE is homogeneous : dy dx = − 2xy x2+3y2 d y d x = − 2 x y x 2 + 3 y 2. The substitution y = vx y = v x leads to. v +x dv dx = −2vx2 x2+3v2x2 = −2v 1+3v2 ⇒ x dv dx = −v − 2v 1 +3v2 = −3v −3v3 1 +3v2 = −3v(1+v2) 1 +3v2 ⇒ 1 +3v2 v(1+v2) dv = −3 dx x v + x d v d x = − 2 v x 2 x 2 + 3 v 2 x 2 = − 2 ...