P8.7Separable differential equations

• B1 for separating the variables correctly, e.g. \(\int \frac{1}{y}\,\mathrm{d}y = \int -4x\,\mathrm{d}x\)
• M1 for integrating both sides, with at least one side correct
• A1 for \(\ln|y| = -2x^{2} + c\)
• A1 for \(y = A\mathrm{e}^{-2x^{2}}\) oe

Worked solution: Separating the variables: \(\int \frac{1}{y}\,\mathrm{d}y = \int -4x\,\mathrm{d}x\).
So \(\ln|y| = -2x^{2} + c\), and hence \(y = A\mathrm{e}^{-2x^{2}}\), where \(A\) is a constant.

• B1 for explaining that \(\mathrm{e}^{x^{2} + c} = \mathrm{e}^{c}\mathrm{e}^{x^{2}}\), not \(\mathrm{e}^{x^{2}} + c\) (the constant multiplies, it is not added)

Worked solution: From \(\ln y = x^{2} + c\), \(y = \mathrm{e}^{x^{2} + c} = \mathrm{e}^{c} \times \mathrm{e}^{x^{2}} = A\mathrm{e}^{x^{2}}\).

• M1 for \(y = A\mathrm{e}^{x^{2}}\) and using \(x = 0\), \(y = 3\)
• A1 for \(y = 3\mathrm{e}^{x^{2}}\)

Worked solution: \(y = A\mathrm{e}^{x^{2}}\); at \(x = 0\), \(3 = A\), so \(y = 3\mathrm{e}^{x^{2}}\).

• B1 for a correct demonstration, e.g. \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\mathrm{e}^{x^{2}}\) but \(2xy = 2x\mathrm{e}^{x^{2}} + 4x\), which differ for \(x \ne 0\)

Worked solution: If \(y = \mathrm{e}^{x^{2}} + 2\) then \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\mathrm{e}^{x^{2}}\), while \(2xy = 2x\mathrm{e}^{x^{2}} + 4x\). These differ by \(4x\), so the equation is not satisfied (except at \(x = 0\)).

• B1 for separating the variables correctly, e.g. \(\int \frac{1}{y^{2}}\,\mathrm{d}y = \int \cos x\,\mathrm{d}x\)
• M1 for integrating both sides, with at least one side correct
• A1 for \(-\frac{1}{y} = \sin x + c\)
• M1 for using the condition \(y = \frac{1}{2}\) when \(x = 0\) to find the constant
• A1 for a correct constant
• A1 for \(y = \frac{1}{2 - \sin x}\) oe

Worked solution: Separating the variables: \(\int \frac{1}{y^{2}}\,\mathrm{d}y = \int \cos x\,\mathrm{d}x\), so \(-\frac{1}{y} = \sin x + c\).
When \(x = 0\), \(y = \frac{1}{2}\): \(-2 = 0 + c\), so \(c = -2\).
So \(y = \frac{1}{2 - \sin x}\).

• M1 for recognising that \(y\) is greatest when \(\sin x\) is greatest, i.e. \(\sin x = 1\)
• A1 for \(1\)

Worked solution: The denominator \(2 - \sin x\) is positive and least when \(\sin x = 1\), so the greatest value of \(y\) is \(\frac{1}{2 - 1} = 1\).