5.18 Differential equationsIB Maths: Analysis and Approaches HL: Revision notes

A differential equation links a function with its derivatives. A first order equation involves only $\frac{dy}{dx}$, for example $\frac{dy}{dx}=x+y$.

• The general solution contains an arbitrary constant: a whole family of curves.
• A particular solution uses an initial condition (e.g. $y=2$ when $x=0$) to fix the constant.

You can always check a solution by differentiating it and substituting back into the equation.

When a differential equation $\frac{dy}{dx}=f(x,y)$ cannot be solved exactly, Euler's method steps along tangent lines with a fixed step length $h$: $y_{n+1}=y_n+h\,f(x_n,y_n),\qquad x_{n+1}=x_n+h.$ For $\frac{dy}{dx}=x+y$, $y(0)=1$, $h=0.1$: $y_1=1+0.1(1)=1.1$, then $y_2=1.1+0.1(0.1+1.1)=1.22$. The exact value is $f(0.2)=2e^{0.2}-1.2\approx1.243$.

If the solution curve is concave up ($\frac{d^{2}y}{dx^{2}}>0$), tangents lie below it and Euler underestimates; concave down gives an overestimate. A smaller $h$ usually gives a better approximation but needs more steps.

If $\frac{dy}{dx}=g(x)h(y)$, rearrange and integrate each side: $\int\frac{1}{h(y)}\,dy=\int g(x)\,dx.$ Example: $\frac{dy}{dx}=3x^{2}y\Rightarrow\ln|y|=x^{3}+c\Rightarrow y=Ae^{x^{3}}$.

The logistic equation $\frac{dn}{dt}=kn(a-n)$ is separable. Use partial fractions: $\frac{1}{n(a-n)}=\frac1a\left(\frac1n+\frac{1}{a-n}\right)$. The solution has the form $n=\frac{a}{1+Be^{-akt}}$, which levels off at the carrying capacity $a$. Growth is fastest when $n=\frac a2$.

A homogeneous differential equation can be written $\frac{dy}{dx}=f\left(\frac yx\right)$, e.g. $\frac{dy}{dx}=\frac{x^{2}+y^{2}}{xy}=\frac xy+\frac yx$.

Substitute $y=vx$, so $\frac{dy}{dx}=v+x\frac{dv}{dx}$ (product rule). The equation becomes separable in $v$ and $x$: $v+x\frac{dv}{dx}=f(v)\;\Rightarrow\;\int\frac{1}{f(v)-v}\,dv=\int\frac1x\,dx.$ Finally replace $v$ by $\frac yx$. For the example with $y(1)=2$: $\frac{v^{2}}{2}=\ln x+2$, so $y=x\sqrt{2\ln x+4}$.

A linear first order equation has the form $\frac{dy}{dx}+P(x)y=Q(x)$. Multiply through by the integrating factor $I(x)=e^{\int P(x)\,dx}.$ The left-hand side then becomes the derivative of a product: $\frac{d}{dx}\left(yI\right)=QI$, so $yI=\int QI\,dx$.

Example: $\frac{dh}{dt}+\frac{h}{t+1}=6$. $I=e^{\ln(t+1)}=t+1$, so $\frac{d}{dt}\left(h(t+1)\right)=6(t+1)$, $h(t+1)=3(t+1)^{2}+C$. With $h(0)=5$, $C=2$ and $h=3(t+1)+\frac{2}{t+1}$.