Standard +0.3 This is a straightforward separable differential equation requiring partial fractions to integrate 1/(3y²+10y+3). The factorization is routine, the separation is immediate, and finding the constant from initial conditions is standard. Slightly easier than average as it follows a well-practiced procedure with no conceptual surprises.
7 Given that \(y = 1\) when \(x = 0\), solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x \left( 3 y ^ { 2 } + 10 y + 3 \right)$$
obtaining an expression for \(y\) in terms of \(x\).
Separate variables and factorise to obtain \(\frac{dy}{(3y+1)(y+3)} = 4x\,dx\) or equivalent
B1
State or imply the form \(\frac{A}{3y+1} + \frac{B}{y+3}\) and use a relevant method to find \(A\) or \(B\)
M1
Obtain \(A = \frac{3}{8}\) and \(B = -\frac{1}{8}\)
A1
Integrate to obtain form \(k_1\ln(3y+1) + k_2\ln(y+3)\)
M1
Obtain correct \(\frac{1}{8}\ln(3y+1) - \frac{1}{8}\ln(y=3) = 2x^2\) or equivalent
A1
Substitute \(x = 0\) and \(y = 1\) in equation of form \(k_1\ln(3y+1) + k_2\ln(y+3) = k_3x^2 + c\) to find a value of \(c\)
M1
Obtain \(c = 0\)
A1
Use correct process to obtain equation without natural logarithm present
M1
Obtain \(y = \frac{3e^{16x^2} - 1}{3 - e^{16x^2}}\) or equivalent
A1
[9]
Separate variables and factorise to obtain $\frac{dy}{(3y+1)(y+3)} = 4x\,dx$ or equivalent | B1 |
State or imply the form $\frac{A}{3y+1} + \frac{B}{y+3}$ and use a relevant method to find $A$ or $B$ | M1 |
Obtain $A = \frac{3}{8}$ and $B = -\frac{1}{8}$ | A1 |
Integrate to obtain form $k_1\ln(3y+1) + k_2\ln(y+3)$ | M1 |
Obtain correct $\frac{1}{8}\ln(3y+1) - \frac{1}{8}\ln(y=3) = 2x^2$ or equivalent | A1 |
Substitute $x = 0$ and $y = 1$ in equation of form $k_1\ln(3y+1) + k_2\ln(y+3) = k_3x^2 + c$ to find a value of $c$ | M1 |
Obtain $c = 0$ | A1 |
Use correct process to obtain equation without natural logarithm present | M1 |
Obtain $y = \frac{3e^{16x^2} - 1}{3 - e^{16x^2}}$ or equivalent | A1 | [9]
7 Given that $y = 1$ when $x = 0$, solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x \left( 3 y ^ { 2 } + 10 y + 3 \right)$$
obtaining an expression for $y$ in terms of $x$.
\hfill \mbox{\textit{CAIE P3 2015 Q7 [9]}}