Standard +0.8 This is a standard second-order linear ODE with constant coefficients requiring both complementary function (solving auxiliary equation with two distinct real roots) and particular integral (trying quadratic form), followed by applying two initial conditions. While methodical with multiple steps for 12 marks, it's a textbook Further Maths exercise with no novel insight required—harder than typical A-level due to being FP2 content but routine within that syllabus.
7. For the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 x ( x + 3 )$$
find the solution for which at \(x = 0 , \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1\) and \(y = 1\).
(Total 12 marks)
2nd M: Substitute their \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) into the D>E> to form an identity in \(x\) with unknown constants
3rd M: Using \(y = 1\) at \(x = 0\) in their general solution to find an equation in \(A\) and \(B\)
4th M: Differentiating their general solution (condone 'slips', but the powers of each term must be correct) and using \(\frac{dy}{dx} = 1\) at \(x = 0\) to find an equation in \(A\) and \(B\)
5th M: Solving simultaneous equations to find both a value of \(A\) and a value of \(B\)
Total: [12]
C.F. $m^2 + 3m + 2 = 0$: $m = -1$ and $m = -2$ | M1 |
$y = Ae^{-x} + Be^{-2x}$ | A12 |
P.I. $y = cx^2 + dx + e$ | B1 |
$\frac{dy}{dx} = 2cx + d$, $\frac{d^2y}{dx^2} = 2c$ | M1 |
$2c + 3(2cx + d) + 2(cx^2 + dx + e) = 2x^2 + 6x$ | |
$2c + 3(2cx + d) + 2(cx^2 + dx + e) = 2x^2 + 6x$ | |
$2c = 2$ | A1 | (One correct value)
$6c + 2d = 6$ | |
$2c + 3d + 2e = 0$ | A1 | (Other two correct values)
$c = 1$ | |
$d = 0$ | |
$e = -1$ | |
General soln: $y = Ae^{-x} + Be^{-2x} + x^2 - 1$ | A1ft5 | (Their C.F. + their P.I.) |
At $x = 0$, $y = 1$: $1 = A + B - 1$ | M1 |
$\frac{dy}{dx} = -Ae^{-x} - 2Be^{-2x} + 2x$, at $x = 0$, $\frac{dy}{dx} = 1$ | M1 |
$1 = -A - 2B$ | |
$(A + B = 2)$ | |
Solving simultaneously: $A = 5$ and $B = -3$ | M1A1 |
Solution: $y = 5e^{-x} - 3e^{-2x} + x^2 - 1$ | A15 |
**1st M:** Attempt to solve auxiliary equation | |
**2nd M:** Substitute their $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the D>E> to form an identity in $x$ with unknown constants | |
**3rd M:** Using $y = 1$ at $x = 0$ in their general solution to find an equation in $A$ and $B$ | |
**4th M:** Differentiating their general solution (condone 'slips', but the powers of each term must be correct) and using $\frac{dy}{dx} = 1$ at $x = 0$ to find an equation in $A$ and $B$ | |
**5th M:** Solving simultaneous equations to find both a value of $A$ and a value of $B$ | |
**Total: [12]**
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