how to solve 2nd order ODE
Answers
Answer:
Step-by-step explanation:
The order of a differential equation is the highest-order derivative that it involves. Thus, a second order differential equation is one in which there is a second derivative but not a third or higher derivative.
To solve a nonhomogeneous linear second-order differential equation, first find the general solution to the complementary equation, then find a particular solution to the nonhomogeneous equation.
If second order difierential equation has the form y = f (t,y ), then the equation for v = y is the first order equation v = f (t,v). Find y solution of the second order nonlinear equation y = −2t (y )2 with initial conditions y(0) = 2, y (0) = −1. + c.
The first major type of second order differential equations you’ll have to learn to solve are ones that can be written for our dependent variable y and independent variable t as:
ad2ydt2+bdydt+cy=0.
Here a, b and c are just constants. In general the coefficients next to our derivatives may not be constant, but fortunately you don’t need to worry about how to approach such problems like that in general for STEP.
Now our approach to solving an equation of the above type is a simple one: we guess a solution. Of course, its an educated guess, there’s a lot of maths behind why we make the guess we do, but essentially it boils down to attempting a solution of the form y=eλt. Here, λ is simply an unknown constant, and our aim becomes to find λ for which a solution of this type satisfies the differential equation. Now, our guess implies that:
dydt=λeλt,d2ydt2=λ2eλt,
and we can therefore substitute into our differential equation to find:
aλ2eλt+bλeλt+ceλt=0.
Now because eλt is never zero, its safe to divide through to acquire a quadratic in λ:
aλ2+bλ+c=0.
This equation, which in future you can jump to straight away, is usually called the Auxiliary Equation.
But we know how to solve quadratics! This means we can find the λ for our a, b and c that allow eλt to satisfy our differential equation. Now, in general we’ll actually have 2 values for λ and our most general solution to the differential equation will be a linear combination of the two solutions they imply. Therefore, if we call our two solutions λ1 and λ2 we have:
y=Aeλ1t+Beλ2t.
But what happens if λ1=λ2? Well then instead we use:
y=(A+Bt)eλ1t.
Additionally, it’s important to realise that our λ may not necessarily be real numbers. If they happen to be complex, we could call our two solutions λ1=r+is and λ2=r−is, since they’ll always be complex conjugate pairs. Then our solution for y, using the relations between eit and the trigonometric functions, can be written as:
y=ert(Acosst+Bsinst).
So these three formula we’ve ended up with are all we actually need to remember! For any homogeneous second order differential equation with constant coefficients, we simply jump to the auxiliary equation, find our (\lambda\), write down the implied solution for y and then use initial conditions to help us find the constants if required.
Inhomogeneous Second Order Differential Equations
One extension to the above that we must tackle is the case when the RHS in our DE is non-zero, i.e. when we have:
ad2ydt2+bdydt+cy=f(t).
Fortunately, the approach is again a simple one. Now, it is common to write our general solution for y in the form y=yc+yp, where yc is known as the complementary function, and yp the particular integral. Specifically, yc is the solution to the problem:
ad2ycdt2+bdycdt+cyc=0,
and yp to:
ad2ypdt2+bdypdt+cyp=f(t).
This may look like we’ve made things infinitely more complicated, but we actually haven’t. Now it should be clear that yc is found from the homogeneous case that we tackled above; so all we need to find it is our auxiliary equation. For yp we again make use of guessing a solution, but our exact guess depends on f. Fortunately there is only a short list of standard guesses you need to remember:
f(t) yp
eαt Peαt
αxn+ lower order powers Pxn+Qxn−1+...+Z
cosαt or sinαt Pcosαt+Qsinαt
In order to find the constants present in yp above, we simpy need to differentiate twice and substitute into its differential equation. Finally, then armed with yc and yp we have our general solution for y and can use initial conditions to find the constants in yc if we require.