Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 11

For a nonzero solution, separation of variables gives $dy/y=kdx/(1+x^2)$, so $\ln |y|=k\arctan x+C$. Restoring the zero solution by allowing an arbitrary real multiplicative constant givesIt is valid for all real numbers $x$, including the constant solutions when $k=0$.

Use integrating factor $e^{-x^3/3}$ for this first-order linear differential equation. Then $(y{e}^{-x^3/3}{)}^{\prime }=x^2{e}^{-x^3/3}=-(e^{-x^3/3}{)}^{\prime }$. Integrating givesThis includes the constant particular solution $y=-1$ when $C=0$.

Use the slope substitution for equations involving x y'-y: put $v=y/x$ on the positive interval containing the initial point. Then $x{y}^{\prime }-y=x^2{v}^{\prime }$, so $x^2{v}^{\prime }=\cot v$. Near $v(1)=\pi /4$, separating variables gives $\tan vdv=dx/x^2$. Integration yields $\ln |\cos v|=1/x+C$. The initial value fixes the sign and constant, givingChoose the inverse cosine branch that agrees with $v(1)=\pi /4$. The solution isOn this interval the argument lies strictly between zero and one and the original cotangent is defined. At the finite lower endpoint, $v\to 0$ and the cotangent becomes singular, so that endpoint is excluded from the maximal interval of the initial value problem. The slope substitution here is valid without classifying the original equation as homogeneous in the usual $y^{\prime }=F(y/x)$ sense.