Jende said:Homework Statement
Find the general solution for the DE: t2y''-2y=0
Homework Equations
These were given for other parts of the problem so I'm not sure if they're relevant.
y1(t)=t2, y2(t)=t-1, y(1)=-2, y'(1)=-7
The Attempt at a Solution
The t2 at the front was really stumping me and I'm not sure where to begin.
Thanks in advance for any help.
LCKurtz said:The earlier parts are relevant. What are ##y_1(t)## and ##y_2(t)##? That is, what do they have to do with this problem? Once you answer that, what do you know about solutions of linear equations?
Jende said:So it should come out to be: y(x)=C1t2+C2t-1
The independent variable is t, not x, so the above should be y(t)=C1t2+C2t-1Jende said:So it should come out to be: y(x)=C1t2+C2t-1
A 2nd order linear differential equation is an equation that involves a function, its derivatives, and constants. The highest derivative in the equation is of second order, meaning it is raised to the power of 2. The equation is also linear, meaning the function and its derivatives are not multiplied together and there are no higher powers of the function or its derivatives.
To find the general solution for a 2nd order linear DE, you must first rearrange the equation so that it is in standard form: y'' + p(x)y' + q(x)y = r(x). Then, you can use methods such as variation of parameters or undetermined coefficients to solve for the general solution. These methods involve finding a particular solution and then adding it to the complementary solution, which is the solution to the associated homogeneous equation.
Sure! Let's take the equation y'' + 2y' + y = x^2. First, we rearrange it to standard form: y'' + 2y' + y - x^2 = 0. Then, we can use the method of undetermined coefficients to find a particular solution. Since the non-homogeneous term is x^2, we can assume a particular solution of the form y_p = Ax^2 + Bx + C. Plugging this into the equation and solving for the coefficients, we get y_p = 1/6x^3 + 1/2x^2 + C. Then, we can find the complementary solution by solving the associated homogeneous equation y'' + 2y' + y = 0. This gives us y_c = Ae^-x + Be^-x. Finally, the general solution is y = y_p + y_c = 1/6x^3 + 1/2x^2 + Ae^-x + Be^-x.
A general solution is a solution that includes all possible solutions to a differential equation, including a particular solution. A particular solution is a specific solution to a differential equation that satisfies the given initial conditions. The general solution can be thought of as a family of solutions, while a particular solution is a single point within that family.
Yes, there are some shortcuts or patterns that can be used for certain types of 2nd order linear DEs. For example, if the non-homogeneous term is a polynomial of the same degree as the highest derivative, you can assume a particular solution of the same form and solve for the coefficients. Additionally, for equations with constant coefficients, there are specific formulas and techniques that can be used to find the general solution. However, it is always important to check your solution to make sure it satisfies the original equation and any given initial conditions.