Differential Equations - Substitutions
tutorial.math.lamar.edu › Classes › DEOct 31, 2019 · Plugging this into the differential equation gives, 1 b(v′ −a) = G(v) v′ = a+bG(v) ⇒ dv a +bG(v) = dx 1 b ( v ′ − a) = G ( v) v ′ = a + b G ( v) ⇒ d v a + b G ( v) = d x. So, with this substitution we’ll be able to rewrite the original differential equation as a new separable differential equation that we can solve.
Substitution Equations - Coping With Calculus
copingwithcalculus.com › substitution-equationsSolve the differential equation using the substitution method: Since e y seems like it might be complicated, let's let v = e y: Now we can make the substitutions for v and dv: Next, we need to try to get this into a form which we can solve: We now have a first-order linear equation in terms of x and v. The first step is finding the integrating factor and multiplying it through the equation: