Solving a Differential Equation by Substitution

When the variables do not separate

$\frac{dy}{dx} = \left(x + y\right)^{2}$ cannot be written as f(x) times g(y), so the variables do not separate as it stands. The question gives a substitution, u = x + y, that turns it into an equation that does.

The first step is to differentiate the substitution with respect to x: $\frac{du}{dx} = 1 + \frac{dy}{dx}$. This links the new derivative to the old one.

Replace every y

From $\frac{du}{dx} = 1 + \frac{dy}{dx}$, $\frac{dy}{dx} = \frac{du}{dx} - 1$. Put that and u = x + y into the equation: $\frac{du}{dx} - 1 = u^{2}$, so $\frac{du}{dx} = 1 + u^{2}$.

Only u and x are left, and the right side depends on u alone, so the variables separate.

Solve, then return to y

Separate and integrate: $\int \frac{1}{{1 + u^{2}}} \,du = \int 1 \,dx$, so $\tan^{-1} u = x + C$ and u = tan(x + C).

The answer must be in x and y, so put x + y back for u: x + y = tan(x + C), and y = tan(x + C) − x.

Check by differentiating: $\frac{dy}{dx} = \sec^{2}\left(x + C\right) - 1 = \tan^{2}\left(x + C\right)$, and tan(x + C) is x + y, so $\frac{dy}{dx} = \left(x + y\right)^{2}$.

The substitution y = vx

$\frac{dy}{dx} = \frac{{x + y}}{x}$ is solved with y = vx, where v is a new variable that depends on x. Since y is a product of v and x, the product rule gives $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substitute: $v + x \frac{dv}{dx} = \frac{{x + \text{vx}}}{x} = 1 + v$, so $x \frac{dv}{dx} = 1$. Separate: $\int 1 \,dv = \int \frac{1}{x} \,dx$, so v = ln|x| + C.

Multiply back by x: y = x ln|x| + Cx. Check: $\frac{dy}{dx} = \ln|x| + 1 + C$, and $\frac{{x + y}}{x} = 1 + \ln|x| + C$.

Two more examples

$\frac{dy}{dx} = \left(2x + y\right)^{2}$ with u = 2x + y. Now $\frac{du}{dx} = 2 + \frac{dy}{dx}$, so $\frac{du}{dx} = 2 + u^{2}$. Integrating gives $\left(\frac{1}{\sqrt{2}}\right) \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) = x + C$, and then $y = \sqrt{2} \tan\left(\sqrt{2}x + D\right) - 2x$, where $D = \sqrt{2}C$.

$\frac{dy}{dx} = \frac{{x^{2} + y^{2}}}{{xy}}$ with y = vx. The right side is $\frac{{1 + v^{2}}}{v} = \frac{1}{v} + v$, so $v + x \frac{dv}{dx} = \frac{1}{v} + v$ and $x \frac{dv}{dx} = \frac{1}{v}$. Then $\int v \,dv = \int \frac{1}{x} \,dx$ gives $\frac{v^{2}}{2} = \ln|x| + C$, and with $v = \frac{y}{x}$, $y^{2} = 2x^{2} \ln|x| + Ax^{2}$.

The usual slips

Treating $\frac{du}{dx}$ as $\frac{dy}{dx}$. For u = 2x + y, $\frac{du}{dx} = 2 + \frac{dy}{dx}$: the 2x term changes with x too.

Writing $\frac{dy}{dx} = \frac{dv}{dx}$ for y = vx. v changes with x, so the product rule gives $v + x \frac{dv}{dx}$.

Stopping at u or v. The question asks for y, so put the substitution back and make y the subject.