Series solution of y''+xy'+y=0, y(0)=1, y'(0)=0 to x^4?

y=1-\frac{x^2}{2}+\frac{x^4}{8}.

What is a reducible differential equation?

One that a given substitution turns into a first-order linear or second-order constant-coefficient equation.

What are the standard types in Core Pure?

First-order linear (integrating factor) and second-order linear with constant coefficients (CF + PI).

With x=e^t, what are \frac{dy}{dx} and \frac{d^2y}{dx^2}?

\frac1x\frac{dy}{dt} and \frac{1}{x^2}\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right).

What must you do after solving the transformed equation?

Substitute back to give the answer in the original variable.

Series solutions and reducible differential equationsEdexcel A-Level Further Maths: Flashcards

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What is the Taylor series method for a differential equation?

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What is the Taylor series method for a differential equation?: Differentiate the equation repeatedly, find $y^{(r)}(0)$ from the initial conditions, and assemble $y=\sum\frac{y^{(r)}(0)}{r!}x^r$ .
For $y''+xy'+y=0$ , write $y''$ and $y'''$ .: $y''=-xy'-y$ and $y'''=-2y'-xy''$ .
Series solution of $y''+xy'+y=0$ , $y(0)=1$ , $y'(0)=0$ to $x^4$ ?: $y=1-\frac{x^2}{2}+\frac{x^4}{8}$ .
Differentiate $y^2$ with respect to $x$ .: $2y\frac{\mathrm{d}y}{\mathrm{d}x}$ (chain rule).
For $y'=x-y^2$ , $y(0)=1$ : what are $y'(0)$ , $y''(0)$ , $y'''(0)$ ?: $-1$ , $3$ , $-8$ .
Series for $y'=x-y^2$ , $y(0)=1$ to $x^3$ ?: $1-x+\frac32x^2-\frac43x^3$ .
What is a reducible differential equation?: One that a given substitution turns into a first-order linear or second-order constant-coefficient equation.
What are the standard types in Core Pure?: First-order linear (integrating factor) and second-order linear with constant coefficients (CF + PI).
Integrating factor for $\frac{\mathrm{d}z}{\mathrm{d}x}+P(x)z=Q(x)$ ?: $\mathrm{e}^{\int P\,\mathrm{d}x}$ .
With $z=\frac1y$ , what is $\frac{\mathrm{d}y}{\mathrm{d}x}$ ?: $-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}$ .
Solve $x y'+y=xy^2$ with $y(1)=1$ .: $y=\frac{1}{x(1-\ln x)}$ .
With $x=\mathrm{e}^t$ , what are $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ?: $\frac1x\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{1}{x^2}\left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2}-\frac{\mathrm{d}y}{\mathrm{d}t}\right)$ .
General solution of $x^2y''-2xy'+2y=4x^3$ ?: $y=Ax+Bx^2+2x^3$ .
What must you do after solving the transformed equation?: Substitute back to give the answer in the original variable.

Exam questions on Series solutions and reducible differential equations

The function $y$ satisfies $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}+x\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=0$ , with $y=1$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}=0$ at $x=0$ .
Use your answer to part (b) to find the value of $\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4}$ at $x=0$ .2 marks
The function $y$ satisfies $\dfrac{\mathrm{d}y}{\mathrm{d}x}=x-y^2$ , with $y=1$ at $x=0$ .
Hence find the first four terms of the series solution of the differential equation in ascending powers of $x$ .2 marks
The differential equation $x\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=xy^2$ is to be solved for $x>0$ using the substitution $z=\dfrac1y$ .
Show that the substitution transforms the equation into $\dfrac{\mathrm{d}z}{\mathrm{d}x}-\dfrac{z}{x}=-1$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).