Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 14 / c / Solution

Differentiate $y^{\prime }=-y+3x$ and use the other equation:Since $3x=y^{\prime }+y$, this becomes $y^{\prime \prime }-3y^{\prime }+2y=0$. Its roots are one and two, so $y=A{e}^t+B{e}^{2t}$. The given data imply $y^{\prime }(0)=-2+3=1$, hence $A+B=2$ and $A+2B=1$. Thus $A=3$, $B=-1$, givingRecover the other dependent variable from $x=(y^{\prime }+y)/3$:These expressions satisfy both original equations and the two initial values.