Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 13

For $x\ne 0$, divide by $x^2$ to obtain $y^{\prime }+y/x=1/x^2$, a first-order linear ordinary differential equation. An integrating factor is $x$ on either interval not crossing zero, since the product rule givesIntegration and division by $x$ give the explicit general solutionThe original equation would read $0=1$ at zero, so no solution can include that point.

Rewrite the linear ordinary differential equation as $y^{\prime }+y=e^{3x}$. Multiplication by its integrating factor $e^x$ gives $(e^{x}y{)}^{\prime }=e^{4x}$. Integrating yieldsThe general solution consists of this particular solution and the homogeneous solution $C{e}^{-x}$.

The original PDF has $8y^3-4x$ in the denominator and $dy/dx$ on the left. Rearrange it into the differential formThe cross partial derivatives of its coefficients are both four, so it is an exact differential form. Integrating its first coefficient in $x$ gives a potential $F=\frac{5}{2}{x}^2+4xy+h(y)$. Matching $F_y=4x-8y^3$ gives $h(y)=-2y^4$. Therefore an implicit general solution isDifferentiating verifies $y^{\prime }=(5x+4y)/(8y^3-4x)$. A solution graph is understood only on intervals where its denominator is nonzero; the printed quotient does not define a derivative when $8y^3=4x$.

For real solutions with $x>0$, the square root requires $y\ge 0$. First retain the solution $y\equiv 0$, which would be lost by dividing by a square root of $y$.

For $y>0$, use the homogeneous first-order differential equation substitution $v=\sqrt{y/x}>0$, so $y=x{v}^2$, $\sqrt{xy}=xv$ and $y^{\prime }=v^2+2xv{v}^{\prime }$. The equation becomesThis is now a separable differential equation. Integration gives $-2/v+2\ln v=-\ln x+C$. Since $x{v}^2=y$, the positive solutions satisfyThe implicit family is sufficient here because only parts (a) and (b) request an explicit expression. It also shows why the nonnegative choice of $v$ matters: $\sqrt{xy}$ denotes the positive square root.