FP2.4 - First order differential equations

A first-order equation is separable when it can be written as $$\frac{dy}{dx}=g(x)h(y),$$ so every factor involving $y$ can be placed with $dy$ and every factor involving $x$ with $dx$. The constant of integration produces a family of solution curves; an initial condition selects one member.

For $$\frac{dy}{dx}=x(1+y),\qquad y(0)=0,$$ first record the equilibrium solution $y=-1$. For the other solutions,$$\int\frac{1}{1+y}\,dy=\int x\,dx,$$so $$\ln|1+y|=\frac{x^2}{2}+C\quad\Longrightarrow\quad y=Ae^{x^2/2}-1.$$The condition $y(0)=0$ gives $A=1$, hence $y=e^{x^2/2}-1$.

For the full family $y=Ae^{x^2/2}-1$, every curve is symmetric about the $y$-axis and has $dy/dx=0$ at $x=0$. Curves with $A>0$ lie above the equilibrium and have a minimum at $x=0$; curves with $A<0$ lie below it and have a maximum there. This qualitative information should agree with any plotted member.

A first-order linear equation must first be written in the standard form $$\frac{dy}{dx}+P(x)y=Q(x).$$ Its integrating factor is $$\mu(x)=e^{\int P(x)\,dx}.$$ Because $\mu'=P\mu$, multiplying by $\mu$ turns the entire left side into the product derivative $\dfrac{d}{dx}(\mu y)$.

For $$\frac{dy}{dx}-2y=x,$$ we have $P=-2$ and $\mu=e^{-2x}$. Thus $$(e^{-2x}y)'=xe^{-2x}.$$Integration by parts gives $$e^{-2x}y=-e^{-2x}\left(\frac{x}{2}+\frac14\right)+C,$$ so $$y=-\frac{x}{2}-\frac14+Ce^{2x}.$$ If $y(0)=\tfrac34$, then $C=1$.

The term containing $C$ is the homogeneous part of the family, while the remaining terms form one particular response to $Q(x)$. Their sum is still a solution because the equation is linear.

A given substitution is useful when it converts a non-standard first-order equation into either a separable equation or a linear equation. The essential move is to differentiate the substitution with the chain rule, replace every occurrence of the old dependent variable consistently, and solve the resulting equation in the new variable.

After differentiating, substitute into the original equation and simplify until only $x$, $v$ and $dv/dx$ remain. Identify the transformed equation as separable or linear, solve it by the corresponding earlier method, restore the integration constant, and finally replace $v$ to express the result in the requested variable and form.

Consider $$\frac{dy}{dx}+2y=xy^3$$ with the given substitution $v=y^{-2}$. Since $$\frac{dv}{dx}=-2y^{-3}\frac{dy}{dx},$$ multiplying the original equation by $-2y^{-3}$ gives $$\frac{dv}{dx}-4v=-2x.$$This is linear. With integrating factor $e^{-4x}$,$$ve^{-4x}=\int-2xe^{-4x}\,dx=e^{-4x}\left(\frac{x}{2}+\frac18\right)+C,$$so $$v=\frac{x}{2}+\frac18+Ce^{4x},\qquad y^2=\frac{1}{\frac{x}{2}+\frac18+Ce^{4x}}.$$