Solving this and substituting back results in a 1 Solving this and substituting back results in a 1-parameter family of implicit solutions: . Getting an explicit solution now is cheap thrills. ______________________________________________________________________ Bonus Noise : The substitution v = y ′ converts the ODE into the first order linear equation (*) Using the integrating factor µ = e x leads to the explicit solution to (*) . Finally, . TEST1/MAP2302 Page 2 of 3 ______________________________________________________________________ (c) Linear as written. Near θ = 0, an integrating factor is easy to come by: for θ ε (- π /2, π /2). Multiplying both sides of the DE by µ results in the following derivative equation: By integrating, we have Applying the initial condition, it follows that C = 4. Hence, an explicit solution near θ = 0 is given by (d) After minimal algebra, this may be seen as either exact with an easy solution, or homogeneous , of degree 1, with a messier solution. As Exact : The usual prestidigitation leads to A 1-parameter family of solutions: As Homogeneous: The degree of homogeneity is 1. Thus, write the equation in the form of dy / dx = g ( y / x ) by doing suitable algebra carefully. After setting y = vx , substituting, and doing a bit more algebra, you will end up looking at the separable equation .
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