2.6 - Substitutions in differential equations | Cambridge A-Level Further Mathematics

A given substitution can turn variable coefficients into constant coefficients, or turn a first-order equation into one with separable variables. You will derive the transformed derivatives, preserve solutions that division might lose, and transfer initial conditions back to the original variables.

Changing x to exponential t

Why do the combinations $x^2y''$ and $xy'$ often appear together? Under the substitution $x=e^t$ , multiplication by $x$ cancels the chain-rule denominator in the first derivative.

Write $Y(t)=y(e^t)$ to distinguish the transformed function from $y(x)$ . Since $\mathrm{d}x/\mathrm{d}t=x$ , the chain rule gives

Y_t=xy',\qquad Y_{tt}=x\frac{\mathrm{d}}{\mathrm{d}x}(xy')=xy'+x^2y''.

The subscripts here mean derivatives with respect to $t$ , while primes still mean derivatives with respect to $x$ .

Derivatives under the given substitution

For $x=e^t$ , so $x>0$ and $t=\ln x$ ,

xy'=Y_t,\qquad x^2y''=Y_{tt}-Y_t.

Consequently

ax^2y''+bxy'+cy=aY_{tt}+(b-a)Y_t+cY.

These transformation formulae are not in the formula booklet: derive them by the chain rule.

The subtraction of $Y_t$ in the second derivative is essential. It is produced by differentiating the extra $x$ , not by a change in the equation's order.

Worked example 1: real roots after substitution

Question. Use $x=e^t$ to solve $x^2y''+xy'-4y=0$ , for $x>0$ .

Plan. Transform both derivative terms, solve the constant-coefficient equation and replace $t$ by $\ln x$ .

(Y_{tt}-Y_t)+Y_t-4Y=0,

so $Y_{tt}-4Y=0$ . Its auxiliary roots are $2,-2$ , giving $Y=Ae^{2t}+Be^{-2t}$ .

Answer. $y=Ax^2+Bx^{-2}$ , for $x>0$ .

Check. For $y=x^r$ , direct differentiation gives the original left-hand side as $(r^2-4)x^r$ . Both $r=2$ and $r=-2$ make it zero, confirming the powers without using the transformed equation.

Change only the sign of the term in $y$ . The transformed roots become complex, so the logarithm appears inside trigonometric functions.

Worked example 2: oscillations in logarithmic x

Question. Use $x=e^t$ to solve $x^2y''+xy'+4y=0$ , for $x>0$ .

Plan. Use the same derivative transformation, then return the trigonometric complementary function to $x$ .

The transformed equation is $Y_{tt}+4Y=0$ , so $Y=A\cos2t+B\sin2t$ .

Answer. $y=A\cos(2\ln x)+B\sin(2\ln x)$ , for $x>0$ .

Check. For $z=\cos(2\ln x)$ ,

z'=-\frac{2\sin(2\ln x)}x,\qquad z''=\frac{2\sin(2\ln x)-4\cos(2\ln x)}{x^2}.

Thus $x^2z''+xz'+4z=0$ . Differentiating the sine term gives the same cancellation, with sine and cosine interchanged.

The substitution $x=e^t$ covers positive $x$ only. Do not claim a negative-domain solution from it without a separately valid change of variable.

Watch out

$x^2y''$ becomes $Y_{tt}-Y_t$ , not $Y_{tt}$ . Derive this line before substituting into the equation.

Use the same cancellation at a new frequency.

Substituting y equals ux

When a first-order equation depends on the ratio $y/x$ , the given substitution $y=ux$ makes that ratio simply $u$ .

Here $u$ is a function of $x$ , not a constant. The product rule gives $y'=u+xu'$ . For $y'=F(y/x)$ , this reduces the equation to

x\frac{\mathrm{d}u}{\mathrm{d}x}=F(u)-u.

It is separable where $x\ne0$ and $F(u)-u\ne0$ . Before dividing by the latter expression, check its zeros: a constant value of $u$ may give an additional straight-line solution $y=ux$ .

Such ratio equations are often called homogeneous first-order equations. That use of “homogeneous” differs from the zero-right-hand-side linear equation used to find a complementary function.

Worked example 3: a logarithmic ratio

Question. Use $y=ux$ to solve $y'=1+y/x$ , for $x>0$ .

Plan. Differentiate the product, cancel the repeated $u$ term, then integrate.

\begin{aligned} u+xu'&=1+u \\ u'&=\frac1x \\ u&=\ln x+C. \end{aligned}

Multiply by $x$ to return to the original dependent variable.

Answer. $y=x(\ln x+C)$ , for $x>0$ .

Check. Direct differentiation gives $y'=\ln x+C+1$ , while $1+y/x=1+\ln x+C$ . The two sides agree for every constant.

Replace the added constant by the square of the ratio. Division now risks losing a solution.

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2.6 - Substitutions in differential equations

Changing x to exponential t

Substituting y equals ux

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