Basic Methods to Solve 1st-Order ODEs - Rhea

By now we have known what is a differential equation and how its solutions conduct. It's time to solve it, like plenty of linear equations we have done before.

3.1 Separable Equation

The easiest method is to separate the variables. This method is switching the variables to make the same variable on the same side, in order to integral on both sides and solve out the function (solution) The standard form of differential equation to use this method is like $ \frac{dy}{dt}=f(y)g(t) $, where $ f(y) $ and $ g(t) $ are easy to be separated out.

For example, we want to solve the differential equation $ \frac{dy}{dt}=-2yt $, with the initial value $ y(0)=1 $.

Now we start separating our variables. We put all items with respect to our dependent variable $ y $ on the left hand side of the equation, and all with respect to our independent variable $ t $ on the right hand side. Hence we get $ \frac{1}{y}dy=-2tdt $.

Integrate on both sides $ \int\frac{1}{y}dy=\int-2tdt $ and get $ ln|y|=-t^2+C $, where $ C $ is a constant.

Reconstruct this equation, $ y=e^{-t^2+C}=e^C e^{-t^2}=Ae^{-t^2} $, where $ A $ is a constant, $ A=e^C $.

Plug in the initial value $ y(0)=1 $. We have $ Ae^0=1 $, so $ A=1 $.

So the final solution is $ y=e^{-t^2} $. This corresponds to the concept we built up in the previous tutorials: The solution to a differential equation is a function.

3.2 Integrating Factor

Those differential equations that can be solved by the above method are pretty special (and easy). There is a more common form for first-ordered differential equations: $ \frac{dy}{dt}+p(t)y=q(t) $, where $ p(t) $ and $ q(t) $ are polynomials with respect to $ t $. Sometimes we have coefficients for $ \frac{dy}{dt} $, just divide them on both sides to obtain the standard form. This method is constructing a "trivial" derivative of a function, then integrating on both sides to obtain the final solution.

For example, we want to solve the differential equation $ t\frac{dy}{dt}=te^{2t}-y $, with the initial value $ y(0.5)=4 $. Divide the coefficient $ t $ of $ \frac{dy}{dt} $, we have $ \frac{dy}{dt}+\frac{1}{t}y=e^{2t} $.

The key part for this method is to find the multiplier, which is known as integrating factor, to construct the derivative structure. The hardworking mathematicians found an effective integrating factor, $ e^{\int p(t)dt} $. In this case, $ p(t)=\frac{1}{t} $. So our integrating factor $ I=e^{\int \frac{1}{t}dt}=e^{ln|t|}=|t| $. From our initial condition, when $ t=0.5 $, $ y=4 $. Hence $ t>0 $. So $ I=t $.

Multiply the factor on both sides of the equation, $ t\frac{dy}{dt}+y=te^{2t} $.

Apply the inverse of product rule for calculating derivative $ fg'+f'g=\frac{d}{dx} $. Replace $ f $ by $ I=t $, and replace $ g $ by $ y $ to obtain $ t\frac{dy}{dt}+y=ty'+t'y=\frac{d}{dt}(ty) $. Hence we have $ \frac{d}{dt}(ty)=te^{2t} $.

Integrate on both sides, $ ty=\int te^{2t}dt=\frac{t}{2}e^{2t}-\frac{1}{4}e^{2t}+C $.

Rearrange the equation to have $ y=\frac{1}{2}e^{2t}-\frac{1}{4t}e^{2t}+\frac{C}{t} $.

Plug in the initial value $ y(0.5)=4 $ to have $ y=\frac{1}{2}e-\frac{1}{2}e+2C=2C=4 $, $ C=2 $.

Therefore, the explicit solution to the differential equation is $ y=\frac{1}{2}e^{2t}-\frac{1}{4t}e^{2t}+\frac{2}{t} $.

3.3 The Exact Differential Equation

An exact differential equation is a certain type of differential equation. So first of all, what does "exact" mean?

Suppose we have a compound function $ \psi (x,y(x))=C $, where $ y $ is a function of $ x $ and $ C $ is a constant. Finding its derivative, we have $ \frac{d}{dx}\psi (x,y(x))=\frac{∂\psi}{∂x}+\frac{∂\psi}{dy} \frac{dy}{dx}=0 $ by chain rule.

By the expansion of Taylor series, $ \psi (x,y(x)) $ can always be written as $ f_1(x)g_1(y)+f_2(x)g_2(y)+...+f_n(x)g_n(y) $. Then still finding its derivative but now by product rule, we have

In Oxford English Dictionary, "exact" in science is defined as "using accurate measurements and following set rules". Here is another intuitive explanation from the perspective of language. You don't need to know the Mandarin language, just kinda feel the pattern of these characters. "Exact" here means "正合", where "正" is formed with all horizontal and vertical character strokes and gives you a feeling of "straightforward, not skewed", and "合" comes with those "straightforward, not skewed" strokes in the bottom and is topped with two strokes going up and finally being together. "Being together" is also what the character "合" basically means.

So this is a perceptual understanding of exact differential equation.

3.5 References

Exact Equations Intuition. Khan Academy. Retrieved from https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/exact-equations/v/exact-equations-intuition-1-proofy on 15 Nov 2017.

Institute of Natural and Mathematical Science, Massey University. (2017). 160.204 Differential Equations I: Course materials. Auckland, New Zealand.