Examples

Homogeneous equation

Homogeneous equation

Consider the differential equation $x{y}^{\prime }=2x+3y$. Assuming $x\ne 0$, this is equivalent to the homogeneous equation $\frac{dy}{dx}=\frac{2x+3y}{x}$. Define $z=y/x$, and we find:

$$x\frac{dz}{dx}+z=2+3z\text{and so}\frac{dz}{1+z}=\frac{2}{x}dx.$$

The solution is therefore $\ln |1+z|=2\ln |x|+c$, or $1+z=C{x}^2$ for any $C\in \mathbb{R}$, i.e. $y=C{x}^3-x$ .

Integrating factor - converting to exact

Integrating factor: converting to exact

Consider the differential equation $(2x^2+y)dx+(x^{2}y-x)dy=0$. This is not exact because $M_y=1$ while $N_x=2xy-1$. However, $\frac{1}{N}(M_y-N_x)=-2/x=:g(x)$ is a function of $x$ alone. Therefore we can use $\mu (x)=e^{\int gdx}=x^{-2}$ as an integrating factor. Multiplying through by $x^{-2}$ produces the equation $(2+y{x}^{-2})dx+(y-x^{-1})dy=0$, which is exact because now $M_y=N_x=x^{-2}$.

Reduction of order

Reduction of order

Consider $x^2{y}^{\prime \prime }=(y^{\prime }{)}^2$. This equation is missing $y$ so we can reduce it to a first-order equation by letting $y^{\prime }=p$ and $y^{\prime \prime }=p^{\prime }$. The equation becomes $x^2{p}^{\prime }=p^2$ which is separable. The solution is $p=\frac{x}{1+c_{1}x}$, and this implies $y=\frac{x}{c_1}-\frac{1}{c_1^2}\ln (1+c_{1}x)+c_2$. (Integrating $p(x)=y^{\prime }(x)$ to obtain $y(x)$.)

Consider $y^{\prime \prime }+k^{2}y=0$. This equation is missing $x$ so we can reduce it to a first-order equation by letting $y^{\prime }=p$ and $y^{\prime \prime }=p^{\prime }p$. The equation becomes $p\frac{dp}{dy}+k^{2}y=0$ which is separable. The solution is $p(y)=\frac{dy}{dx}=\pm k\sqrt{c_1-y^2}$. This is a new differential equation which is also separable. Its general solution is $y=\sqrt{c_1}\sin (\pm kx+c_2)$.

Swapping roles

Swapping roles

Consider the differential equation $y^{\prime }{y}^2=y+x{y}^{\prime }$. If we change $y^{\prime }$ to $1/x^{\prime }$, the equation becomes

$$\frac{y^2}{x^{\prime }}=y+\frac{x}{x^{\prime }}$$

or $x^{\prime }y+x=y^2$ or $x^{\prime }+(1/y)x=y^2$. (This is linear of the form $x^{\prime }+a(y)x=b(y)$.) To solve it we write $(xy{)}^{\prime }=y^2$ and thus $xy=\frac{y^3}{3}+C$ or $x=\frac{y^2}{3}+\frac{C}{y}$.