6. (a) Find the general solution of the differential equation
$$6 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 6 y = x - 6 x ^ { 2 }$$
(b) Find the particular solution for which \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 2 }\) when \(x = 0\)
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Question 6:
Part (a):
Answer Marks
Guidance
Working Mark
Notes
\(6m^2 + 5m - 6 = 0 \Rightarrow (3m-2)(2m+3) = 0\) M1
Forms and solves auxiliary equation
\(m = \frac{2}{3}, \frac{-3}{2}\) A1
Correct roots
CF: \(Ae^{\frac{2}{3}x} + Be^{-\frac{3}{2}x}\) B1ft
CF of form shown using their 2 real roots; can be awarded if seen in general solution
PI: \((y=) Cx^2 + Dx + E\) B1
May include higher powers
\(\frac{dy}{dx} = 2Cx + D,\ \frac{d^2y}{dx^2} = 2C\) M1
Differentiates PI twice; all powers of \(x\) to decrease by 1
\(6(2C) + 5(2Cx+D) - 6(Cx^2+Dx+E) \equiv -6x^2+x\) M1
Substitutes derivatives into equation and equates at least one pair of coefficients
\(-6C = -6\); \(10C - 6D = 1\); \(12C + 5D - 6E = 0\)
\(C=1\); \(D = \frac{3}{2}\); \(E = \frac{13}{4}\) M1
Attempt to solve 3 equations; must reach numerical value for all 3 coefficients
\(y = Ae^{\frac{2}{3}x} + Be^{-\frac{3}{2}x} + x^2 + \frac{3}{2}x + \frac{13}{4}\) A1
Must start \(y = \ldots\); cao
(8 marks)
Part (b):
Answer Marks
Guidance
Working Mark
Notes
\(\frac{dy}{dx} = \frac{2}{3}Ae^{\frac{2}{3}x} - \frac{3}{2}Be^{-\frac{3}{2}x} + 2x + \frac{3}{2}\) M1
Differentiates their GS – min 4 terms
\(y=0,\ \frac{dy}{dx}=\frac{3}{2},\ x=0\): \(0 = A+B+\frac{13}{4}\); \(\frac{3}{2} = \frac{2}{3}A - \frac{3}{2}B + \frac{3}{2}\) M1
Forms 2 simultaneous equations using given boundary values
\(4A + 4B = -13,\ 4A - 9B = 0\) M1
Attempt to solve; must reach \(A=\ldots\) or \(B=\ldots\)
\(A = -\frac{9}{4},\ B = -1\) A1
Both correct
\(y = x^2 + \frac{3}{2}x + \frac{13}{4} - \frac{9}{4}e^{\frac{2}{3}x} - e^{-\frac{3}{2}x}\) A1
Must start \(y = \ldots\)
(5 marks) — Total: 13
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# Question 6:
## Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $6m^2 + 5m - 6 = 0 \Rightarrow (3m-2)(2m+3) = 0$ | M1 | Forms and solves auxiliary equation |
| $m = \frac{2}{3}, \frac{-3}{2}$ | A1 | Correct roots |
| CF: $Ae^{\frac{2}{3}x} + Be^{-\frac{3}{2}x}$ | B1ft | CF of form shown using their 2 real roots; can be awarded if seen in general solution |
| PI: $(y=) Cx^2 + Dx + E$ | B1 | May include higher powers |
| $\frac{dy}{dx} = 2Cx + D,\ \frac{d^2y}{dx^2} = 2C$ | M1 | Differentiates PI twice; all powers of $x$ to decrease by 1 |
| $6(2C) + 5(2Cx+D) - 6(Cx^2+Dx+E) \equiv -6x^2+x$ | M1 | Substitutes derivatives into equation and equates at least one pair of coefficients |
| $-6C = -6$; $10C - 6D = 1$; $12C + 5D - 6E = 0$ | | |
| $C=1$; $D = \frac{3}{2}$; $E = \frac{13}{4}$ | M1 | Attempt to solve 3 equations; must reach numerical value for all 3 coefficients |
| $y = Ae^{\frac{2}{3}x} + Be^{-\frac{3}{2}x} + x^2 + \frac{3}{2}x + \frac{13}{4}$ | A1 | Must start $y = \ldots$; cao |
**(8 marks)**
## Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{dy}{dx} = \frac{2}{3}Ae^{\frac{2}{3}x} - \frac{3}{2}Be^{-\frac{3}{2}x} + 2x + \frac{3}{2}$ | M1 | Differentiates their GS – min 4 terms |
| $y=0,\ \frac{dy}{dx}=\frac{3}{2},\ x=0$: $0 = A+B+\frac{13}{4}$; $\frac{3}{2} = \frac{2}{3}A - \frac{3}{2}B + \frac{3}{2}$ | M1 | Forms 2 simultaneous equations using given boundary values |
| $4A + 4B = -13,\ 4A - 9B = 0$ | M1 | Attempt to solve; must reach $A=\ldots$ or $B=\ldots$ |
| $A = -\frac{9}{4},\ B = -1$ | A1 | Both correct |
| $y = x^2 + \frac{3}{2}x + \frac{13}{4} - \frac{9}{4}e^{\frac{2}{3}x} - e^{-\frac{3}{2}x}$ | A1 | Must start $y = \ldots$ |
**(5 marks) — Total: 13**
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6. (a) Find the general solution of the differential equation
$$6 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 6 y = x - 6 x ^ { 2 }$$
(b) Find the particular solution for which $y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 2 }$ when $x = 0$\\
\hfill \mbox{\textit{Edexcel FP2 2018 Q6 [13]}}