Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 16 (a)/Solution 1

Re-arranging the differential equation yields

$${\begin{aligned}{\frac {\mathrm {d} y}{\mathrm {d} x}}&=2xy-4x-4y+8=2(x-2)(y-2).\end{aligned}}$$

Separating,

$${\begin{aligned}{\frac {1}{y-2}}\;\mathrm {d} y&=(2x-4)\;\mathrm {d} x.\end{aligned}}$$

Taking the integral,

$${\begin{aligned}\int {\frac {1}{y-2}}\;\mathrm {d} y&=\int (2x-4)\;\mathrm {d} x.\\\log |y-2|&=x^{2}-4x+c.\\|y-2|&=e^{x^{2}-4x+c}=Ce^{x^{2}-4x},\end{aligned}}$$

where $C=e^{c}$.

To get rid of the absolute value, note that if $y-2$ is negative, then we can replace $C$ with $-C$. Therefore, for some constant $C$, we have

$${\begin{aligned}y-2=Ce^{x^{2}-4x},\end{aligned}}$$

which simplifies to

$${\begin{aligned}y=2+Ce^{x^{2}-4x}.\end{aligned}}$$