The calculator will try to find the solution of the given ODE: first-order, second-order, nth-order, separable, linear, exact, Bernoulli, homogeneous, or inhomogeneous. Show Initial conditions are also supported. Enter an equation (and, optionally, the initial conditions): For example, y''(x)+25y(x)=0, y(0)=1, y'(0)=2. Write `y'(x)` instead of `(dy)/(dx)`, `y''(x)` instead of `(d^2y)/(dx^2)`, etc. If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below. Solving linear differential equations with constant coefficients reduces to an algebraic problem. There is no similar procedure for solving linear differential equations with variable coefficients. With the exception of special types, such as the Cauchy equations, these will generally require the use of the power series techniques for a solution. Application DetailsPublish Date: May 23,
2007 Power Series Calculator is a free online tool that displays the infinite series of the given function. BYJU’S online power series calculator tool makes the calculation faster, and it displays the expanded form of a given function in a fraction of seconds. How to Use the Power Series Calculator?The procedure to use the power series calculator is as follows: What is Meant by Power Series?In Mathematics, a power series is defined as an infinite series which is similar to the polynomial with many terms. Usually, the power series will converge at a value “x” within a certain interval, such that the absolute value of x is less than some positive value
“r”, which is known as the radius of convergence. \(\begin{array}{l}a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+…\end{array} \) Examples for A differential equation is an equation involving a function and its derivatives. It can be referred to as an ordinary differential equation (ODE) or a partial differential equation (PDE) depending
on whether or not partial derivatives are involved. Wolfram|Alpha can solve many problems under this important branch of mathematics, including solving ODEs, finding an ODE a function satisfies and solving an ODE using a slew of numerical methods. Solve an ODE or find an ODE a function satisfies. y'' + y = 0 w"(x)+w'(x)+w(x)=0 y'' + y = 0, y(0)=2, y'(0)=1 y''(t) + y(t) = sin t
x^2 y''' - 2 y' = x y'(t) = a t y(t) f'(t) = f(t)^2 + 1 y"(z) + sin(y(z)) = 0 differential equations sin 2x differential equations J_2(x) Numerically solve a differential equation using a variety of classical methods. Solve an ODE using a specified numerical method:Runge-Kutta method, dy/dx = -2xy, y(0) = 2, from 1 to 3, h = .25 {y'(x) = -2 y, y(0)=1} from 0 to 2 by implicit midpoint Specify an adaptive method:solve {y'(x) = -2 y, y(0)=1} from 0 to 10 using r k f algorithm GO FURTHERStep-by-Step Solutions for Differential Equations RELATED EXAMPLES$\begingroup$ The equation is $$y'' - xy' + y = 0$$ So far I have the recurrence relation - $$a_{n+2} = \dfrac{(n-1)a_n}{(n+1)(n+2)} $$ From this - $a_2 = \dfrac{-a_0}{2!}$ $a_3 = 0$ $a_4 = \dfrac{-a_0}{4!}$ $a_5 = 0$ $a_6 = \dfrac{-3a_0}{6!}$ and so on.. The question asks for the first five non-zero terms of a general series solution of the d.e, seperating out for $a_0$ and $a_1$ How do I compute this? Thanks asked Apr 27, 2014 at 21:49
$\endgroup$ 11 $\begingroup$ The differential equation \begin{align} y^{''} - x y^{'} + y = 0 \end{align} can be solved via a power series of the form \begin{align} y(x) = \sum_{k=0}^{\infty} a_{n} x^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \cdots . \end{align} It is fairly evident that \begin{align} \sum_{k=0}^{\infty} k(k-1) a_{k} x^{k} = \sum_{k=0}^{\infty} (k-1) a_{k} x^{k} \end{align} which yields the equation for the coefficients \begin{align} a_{k+2} = \frac{ (k-1) a_{k} }{ (k+1) (k+2) }. \end{align} It is discovered that $a_{3} = 0 \cdot a_{1}$. Since, for $k$ being odd, say $k \rightarrow 2k+1$, \begin{align} a_{2k+3} = \frac{k a_{2k+1} }{(k+1)(2k+3)} \end{align} it is clear that all the odd coefficients depend of $a_{3}$ for $k \geq 1$ and leads to $a_{2k+1} = 0$ for $k \geq 1$. The even $k$ values are \begin{align} a_{2} &= - \frac{a_{0}}{2!} \\ a_{4} &= - \frac{a_{0}}{4!} \\ a_{6} &= - \frac{(1 \cdot 3) a_{0}}{6!} \\ a_{8} &= - \frac{(1\cdot 3 \cdot 5)a_{0}}{8!} \end{align} which has the general form \begin{align} a_{2k} = - \frac{a_{0}}{2^{k} k! (2k-1)}. \end{align} The series for $y(x)$ now be seen in the form \begin{align} y(x) = a_{0} + a_{1} x - a_{0} \sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)}. \end{align} The power series discovered can be evaluated as follows. Consider \begin{align} \partial_{x} \left( \sum_{k=1}^{\infty} \frac{ x^{2n-1} }{2^{k} k! (2k-1)} \right) &= \sum_{k=1}^{\infty} \frac{ x^{2n-2} }{2^{k} k!} = \frac{1}{x^{2}}( e^{x^{2}/2} -1). \end{align} Integrating both sides \begin{align} \sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)} &= x \int^{x} \frac{e^{u^{2}/2} -1}{u^{2}} du = x \left[ \sqrt{\frac{\pi}{2} } erfi\left( \frac{x}{\sqrt{2}} \right) - \frac{e^{x^{2}/2}}{x} + \frac{1}{x} \right] \\ &= \sqrt{\frac{\pi}{2} } \cdot x \cdot erfi\left( \frac{x}{\sqrt{2}} \right) - e^{x^{2}/2} + 1. \end{align} With this series the general solution of $y(x)$ can be sen by \begin{align} y(x) &= a_{0} + a_{1} x - \sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)} \\ &= a_{1} x - a_{0} \left[ \sqrt{\frac{\pi}{2} } \cdot x \cdot erfi\left( \frac{x}{\sqrt{2}} \right) - e^{x^{2}/2} \right], \end{align} where $erfi(x)$ is the imaginary error function (erfi(x) = -i erf(ix)). answered May 9, 2014 at 5:20
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Power series solution of differential equations calculator
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