Pc11, Pc12 - Second order homogeneous equations | OCR A-Level Further Mathematics B (MEI)

A quadratic equation can determine the whole family of solutions of a second-order differential equation. Pc11 uses this auxiliary equation; Pc12 explains how distinct real, repeated real and complex roots lead to different solution forms. You will choose the correct form and use two conditions to determine its two constants.

Distinct real auxiliary roots

How can the equation $y''+2y'+5y=0$ produce a curve that crosses zero repeatedly, while other second-order equations give only exponential shapes?

The clue is the shape of a function under differentiation. If $y=e^{rx}$ , then $y'=re^{rx}$ and $y''=r^2e^{rx}$ . Substitution into $y''+ay'+by=0$ gives $(r^2+ar+b)e^{rx}=0$ . The exponential is never zero, so a quadratic in $r$ decides which exponential solutions are possible.

Homogeneous equation and auxiliary equation

A linear equation $y''+ay'+by=0$ , with constant real $a$ and $b$ , is homogeneous because its right side is zero. Its auxiliary equation is $r^2+ar+b=0$ . The homogeneous general solution is also called the complementary function. A non-zero right side, such as $y''+ay'+by=x$ , is not homogeneous.

If a leading coefficient other than $1$ occurs, divide through first or retain that coefficient in the quadratic consistently. The equation's order is two, so its general solution needs two independent constants; two supplied conditions can select one particular solution.

Two distinct real roots

For constant-coefficient $y''+ay'+by=0$ , if $a^2-4b>0$ , the auxiliary equation has distinct real roots $r_1,r_2$ , and

y=Ae^{r_1x}+Be^{r_2x}

where $A$ and $B$ are arbitrary real constants. A zero root contributes a constant term, since $e^{0x}=1$ .

Not in the formula booklet: learn it.

The sum works because differentiating and adding are linear operations: each term separately satisfies the equation, so their linear combination does too. Conditions must be applied to the entire sum.

Worked example 1: two growing exponentials

Question. Solve $y''-3y'+2y=0$ , given $y(0)=2$ and $y'(0)=3$ .

Plan. Solve the auxiliary quadratic, form the general solution, then use the value and derivative conditions.

\begin{aligned} r^2-3r+2&=(r-1)(r-2)=0 && \text{auxiliary equation} \\ r&=1,2 && \text{distinct roots} \\ y&=Ae^x+Be^{2x} && \text{general solution} \\ y'&=Ae^x+2Be^{2x} && \text{differentiate} \\ A+B&=2,\quad A+2B=3 && \text{conditions at zero} \\ B&=1,\quad A=1 && \text{solve simultaneously} \end{aligned}

Why this step? The derivative condition gives a different equation in the same constants. Using $y'(0)=3$ as if it were another value of $y$ would lose the information about the curve's initial slope.

Answer. $y=e^x+e^{2x}$ .

Check. At $0$ , $y=2$ and $y'=3$ . In the differential equation, the $e^x$ coefficient is $1-3+2=0$ , and the $e^{2x}$ coefficient is $4-6+2=0$ .

The next example changes only the coefficient of $y$ , from $2$ to $-4$ . The initial conditions are unchanged; one auxiliary root is now negative.

Worked example 2: one growing and one decaying term

Question. Solve $y''-3y'-4y=0$ , given $y(0)=2$ and $y'(0)=3$ .

Plan. Find the new roots before choosing the exponential terms, then apply the same two conditions.

\begin{aligned} r^2-3r-4&=(r-4)(r+1)=0 && \text{auxiliary equation} \\ r&=4,-1 && \text{new roots} \\ y&=Ae^{4x}+Be^{-x} && \text{general solution} \\ y'&=4Ae^{4x}-Be^{-x} && \text{differentiate} \\ A+B&=2,\quad4A-B=3 && \text{initial conditions} \\ 5A&=5\implies A=1,\quad B=1 && \text{add the equations} \end{aligned}

Why this step? The root determines the complete exponent, including its sign. A negative root gives a decaying term as $x$ increases.

Answer. $y=e^{4x}+e^{-x}$ .

Check. The initial values are $2$ and $4-1=3$ . Each exponential's coefficient in $y''-3y'-4y$ is zero because its exponent satisfies the auxiliary equation.

The general solution contains arbitrary constants even when a particular solution later turns one of them into zero. Do not choose constants before using the given conditions.

Watch out

Wrong: replace a root $r=2$ by a term $2e^x$ . Right: it gives $Be^{2x}$ ; the root belongs in the exponent, while the coefficient remains an arbitrary constant.

The following quadratic has new roots. Use both initial conditions after forming the general family.

A repeated root needs a second shape

A repeated root does not give two independent copies of the same exponential; multiplying one copy by $x$ supplies the missing solution shape.

Writing $Ae^{rx}+Be^{rx}$ merely gives $(A+B)e^{rx}$ , with only one effective constant. When the root is repeated, $xe^{rx}$ also satisfies the equation and is not a constant multiple of $e^{rx}$ . This makes the two-constant family possible.

A repeated real root

If $a^2-4b=0$ , the auxiliary equation has repeated root $r=-a/2$ , and

y=(A+Bx)e^{rx}

For the repeated root $r=0$ , this becomes $y=A+Bx$ .

Not in the formula booklet: learn it.

Differentiate the whole product before applying a slope condition. The derivative of $A+Bx$ contributes a term that is easy to miss.

Worked example 3: a repeated negative root

Question. Solve $y''+4y'+4y=0$ , given $y(0)=1$ , $y'(0)=0$ .

Plan. Use the repeated-root form, differentiate by the product rule and apply the conditions.

\begin{aligned} r^2+4r+4&=(r+2)^2=0 && \text{repeated root }-2 \\ y&=(A+Bx)e^{-2x} && \text{general solution} \\ y'&=\bigl(B-2A-2Bx\bigr)e^{-2x} && \text{product rule} \\ A&=1,\quad B-2A=0 && \text{initial conditions} \\ B&=2 && \text{solve} \end{aligned}

Why this step? Differentiating only the exponential would omit the $B$ term. That would make the slope condition impossible to apply correctly.

Answer. $y=(1+2x)e^{-2x}$ .

Check. $y'=-4xe^{-2x}$ , so $y'(0)=0$ . Also $y''=(8x-4)e^{-2x}$ , and substitution gives zero in the differential equation.

The next example changes only the sign of the $y'$ coefficient. The repeated root becomes positive, while the same initial slope requires a different coefficient of $x$ .

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Pc11, Pc12 - Second order homogeneous equations

Distinct real auxiliary roots

A repeated root needs a second shape

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