| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (polynomial/exponential x-side) |
| Difficulty | Moderate -0.3 This is a straightforward separable variables question with scaffolding. Part (a) directly provides the integral needed for part (b). The separation is simple (divide by y^{1/2}), and the integration is routine. Slightly easier than average due to the helpful scaffolding, but still requires proper technique for separable DEs. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{4x+3}{x}\,dx \rightarrow \int \ldots + \frac{b}{x}\,dx = \ldots + \ldots\ln x\) | M1 | Attempts to divide to obtain \(\ldots + \frac{b}{x}\) and uses \(\int \frac{1}{x}\,dx = \ln x\) or \(\int \frac{1}{x}\,dx = \ln kx\) |
| \(= 4x + 3\ln x + (c)\) | A1 | No requirement for \(+ c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{4x+3}{x}\,dx = \int(4x+3)x^{-1}\,dx = (4x+3)\ln x - \int 4\ln x\,dx\) \(= (4x+3)\ln x - 4x\ln x + kx\) | M1 | Requires 2 applications of parts to obtain expression of this form |
| \(= (4x+3)\ln x - 4x\ln x + 4x\,(+c)\) | A1 | No requirement for \(+ c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{4x+3}{x}\,dx = \int(4x+3)x^{-1}\,dx = (2x^2+3x)x^{-1} + \int(2x^2+3x)x^{-2}\,dx\) \(= (2x+3) + \int\left(2 + 3x^{-1}\right)dx = 2x+3+2x+3\ln x\,(+c)\) | M1 | Requires applications of parts to obtain expression of this form |
| \(= (2x^2+3x)x^{-1} + 2x + 3\ln x\,(+c)\) | A1 | No requirement for \(+ c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{(4x+3)y^{\frac{1}{2}}}{x} \Rightarrow \int \frac{1}{y^{\frac{1}{2}}}\,dy = \int \frac{(4x+3)}{x}\,dx\) | B1 | Separates variables correctly. Accept with or without integral signs, possibly without "\(dx\)" and/or "\(dy\)". Look for \(\frac{1}{y^{\frac{1}{2}}} = \frac{(4x+3)}{x}\) |
| \(2y^{\frac{1}{2}} = 4x + 3\ln x + c\) | M1 | Look for \(ky^{\frac{1}{2}} =\) their (a), or \(ky^{\frac{1}{2}} =\) an attempt at \(\int \frac{4x+3}{x}\,dx\) |
| \(2y^{\frac{1}{2}} = 4x + 3\ln x + c\) | A1 | Correct equation including \(+ c\) |
| \(x = 1,\ y = 25 \Rightarrow 2(25)^{\frac{1}{2}} = 4(1) + 3\ln(1) + c \Rightarrow c = \ldots\) | M1 | Substitutes \(x = 1\) and \(y = 25\) into integrated equation and proceeds to obtain a value for \(c\) |
| \(y = \left(2x + \frac{3}{2}\ln x + 3\right)^2\) | A1 | Correct equation including "\(y =\)". The \(2x + \frac{3}{2}\ln x + 3\) can be in any equivalent correct form. |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{4x+3}{x}\,dx \rightarrow \int \ldots + \frac{b}{x}\,dx = \ldots + \ldots\ln x$ | M1 | Attempts to divide to obtain $\ldots + \frac{b}{x}$ and uses $\int \frac{1}{x}\,dx = \ln x$ or $\int \frac{1}{x}\,dx = \ln kx$ |
| $= 4x + 3\ln x + (c)$ | A1 | No requirement for $+ c$ |
## Part (a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{4x+3}{x}\,dx = \int(4x+3)x^{-1}\,dx = (4x+3)\ln x - \int 4\ln x\,dx$ $= (4x+3)\ln x - 4x\ln x + kx$ | M1 | Requires 2 applications of parts to obtain expression of this form |
| $= (4x+3)\ln x - 4x\ln x + 4x\,(+c)$ | A1 | No requirement for $+ c$ |
## Part (a) Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{4x+3}{x}\,dx = \int(4x+3)x^{-1}\,dx = (2x^2+3x)x^{-1} + \int(2x^2+3x)x^{-2}\,dx$ $= (2x+3) + \int\left(2 + 3x^{-1}\right)dx = 2x+3+2x+3\ln x\,(+c)$ | M1 | Requires applications of parts to obtain expression of this form |
| $= (2x^2+3x)x^{-1} + 2x + 3\ln x\,(+c)$ | A1 | No requirement for $+ c$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(4x+3)y^{\frac{1}{2}}}{x} \Rightarrow \int \frac{1}{y^{\frac{1}{2}}}\,dy = \int \frac{(4x+3)}{x}\,dx$ | B1 | Separates variables correctly. Accept with or without integral signs, possibly without "$dx$" and/or "$dy$". Look for $\frac{1}{y^{\frac{1}{2}}} = \frac{(4x+3)}{x}$ |
| $2y^{\frac{1}{2}} = 4x + 3\ln x + c$ | M1 | Look for $ky^{\frac{1}{2}} =$ their (a), or $ky^{\frac{1}{2}} =$ an attempt at $\int \frac{4x+3}{x}\,dx$ |
| $2y^{\frac{1}{2}} = 4x + 3\ln x + c$ | A1 | Correct equation including $+ c$ |
| $x = 1,\ y = 25 \Rightarrow 2(25)^{\frac{1}{2}} = 4(1) + 3\ln(1) + c \Rightarrow c = \ldots$ | M1 | Substitutes $x = 1$ and $y = 25$ into integrated equation and proceeds to obtain a value for $c$ |
| $y = \left(2x + \frac{3}{2}\ln x + 3\right)^2$ | A1 | Correct equation including "$y =$". The $2x + \frac{3}{2}\ln x + 3$ can be in any equivalent correct form. |
---2. (a) Find $\int \frac { 4 x + 3 } { x } \mathrm {~d} x , \quad x > 0$\\
(b) Given that $y = 25$ at $x = 1$, solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + 3 ) y ^ { \frac { 1 } { 2 } } } { x } \quad x > 0 , y > 0$$
giving your answer in the form $y = [ \mathrm { g } ( x ) ] ^ { 2 }$.\\
VJYV SIHI NITIIYIM ION OC\\
VI4V SIHI NI JAHMA ION OC\\
VEYV SIHI NI JIIIM ION OO
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\hfill \mbox{\textit{Edexcel C34 2019 Q2 [7]}}