Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 15

Because the equation depends on the ratio $y/x$, set $u=y/x$, so $y^{\prime }=u+x{u}^{\prime }$. Dividing by $x^2$ and substituting givesThis reduces the homogeneous first-order differential equation to a separable differential equation. For $u\ne 1$,At $x=e$, the condition $y=0$ gives $u=0$, so $1=1+C$ and $C=0$. HenceThe interval displayed is the maximal solution interval containing $x=e$: there is a pole at $x=1$. Separating variables omitted the solution $u=1$, or $y=x$, but that solution does not satisfy the given condition.

Differentiate $v^{\prime }=u$ once, then substitute the other equation. The required second-order linear differential equation is

The characteristic equation of the homogeneous equation is ${\lambda }^2-4=0$, giving exponential solutions $e^{2t}$ and $e^{-2t}$. A constant particular solution is $v_p=-1/2$. Thus the general solution is

The equation $u=v^{\prime }$ fixes $u$ from the same two constants:Indeed $u^{\prime }=4A{e}^{2t}+4B{e}^{-2t}=4v+2$, so both original equations hold. No further independent integration constant is available in this two-dimensional linear system of ordinary differential equations.

The dependent variable itself does not occur, so reduce order by putting $p=y^{\prime }$. Then $p^{\prime }+2p=3x$, a first-order linear differential equation. Its integrating factor is $e^{2x}$:Integrating once more and renaming the constants givesThe two homogeneous terms are the constant solution and the exponential associated with the roots $0,-2$ of the characteristic equation.

For the general solution in part (i),Thus $B=-3/8$ and $A=11/8$, giving the solution of the initial value problemDirect substitution checks both the differential equation and the two initial conditions.