Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 6

For the suggested substitution, put $y=xv(x)$. Then the product rule gives $y^{\prime }=v+x{v}^{\prime }$, while the right side of the ordinary differential equation becomes $v+v^{-3}$. Cancelling $v$ leaves the separable differential equationIntegration yields $v^4/4=\log x+C$. The initial condition gives $v(1)=1$, so $v^4=1+4\log x$. The positive root is forced by the initial value and continuity:This is the maximal connected interval containing $1$ on which the real solution exists. At the left endpoint $y=0$, where the original ordinary differential equation is undefined; extending the fourth-root formula further left would also cease to be real.

Use the substitution $u=x^2$, so $du=2xdx$. The antiderivative isDifferentiation by the chain rule gives $4x{e}^{x^2}$, verifying the result.

Divide by $x^2$, which is allowed because $x>0$:The integrating factor for this linear ordinary differential equation is $e^{\int dx/x}=x$. Multiplying through makes the left side a product rule derivative:Part (b) now gives $xy=2e^{x^2}+C$. Therefore

For the homogeneous linear ordinary differential equation, the characteristic equation isIts real solution is $y_h=e^{-x}(C_1\cos x+C_2\sin x)$. For a particular solution to the full ordinary differential equation, try $y_p=ax+b$. Substitution givesso $a=4$ and $b=-2$. Adding the homogeneous and particular solutions yieldsThe two arbitrary real constants account for the second order of the ordinary differential equation.