Solving a Differential Equation by Substitution

When the variables do not separate

$\frac{dy}{dx}={\left(x+y\right)}^2$dy/dx = (x + y)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: $\frac{du}{dx}=1+\frac{dy}{dx}$du/dx = 1 + dy/dx. This links the new derivative to the old one.

Replace every y

From $\frac{du}{dx}=1+\frac{dy}{dx}$du/dx = 1 + dy/dx, $\frac{dy}{dx}=\frac{du}{dx}-1$dy/dx = du/dx − 1. Put that and u = x + y into the equation: $\frac{du}{dx}-1=u^2$du/dx − 1 = u², so $\frac{du}{dx}=1+u^2$du/dx = 1 + u².

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: $\int \frac{1}{1+u^2}du=\int 1dx$∫ 1/(1 + u²) du = ∫ 1 dx, so ${\tan }^{-1}u=x+C$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: $\frac{dy}{dx}={\sec }^2\left(x+C\right)-1={\tan }^2\left(x+C\right)$dy/dx = sec²(x + C) − 1 = tan²(x + C), and tan(x + C) is x + y, so $\frac{dy}{dx}={\left(x+y\right)}^2$dy/dx = (x + y)².

The substitution y = vx

$\frac{dy}{dx}=\frac{x+y}{x}$dy/dx = (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 $\frac{dy}{dx}=v+x\frac{dv}{dx}$dy/dx = v + x dv/dx.

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

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

Two more examples

$\frac{dy}{dx}={\left(2x+y\right)}^2$dy/dx = (2x + y)² with u = 2x + y. Now $\frac{du}{dx}=2+\frac{dy}{dx}$du/dx = 2 + dy/dx, so $\frac{du}{dx}=2+u^2$du/dx = 2 + u². Integrating gives $\left(\frac{1}{\sqrt{2}}\right){\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)=x+C$(1/√2) tan⁻¹(u/√2) = x + C, and then $y=\sqrt{2}\tan \left(\sqrt{2}x+D\right)-2x$y = √2 tan(√2x + D) − 2x, where $D=\sqrt{2}C$D = √2C.

$\frac{dy}{dx}=\frac{x^2+y^2}{xy}$dy/dx = (x² + y²)/(xy) with y = vx. The right side is $\frac{1+v^2}{v}=\frac{1}{v}+v$(1 + v²)/v = 1/v + v, so $v+x\frac{dv}{dx}=\frac{1}{v}+v$v + x dv/dx = 1/v + v and $x\frac{dv}{dx}=\frac{1}{v}$x dv/dx = 1/v. Then $\int vdv=\int \frac{1}{x}dx$∫ v dv = ∫ 1/x dx gives $\frac{v^2}{2}=\ln |x|+C$v²/2 = ln|x| + C, and with $v=\frac{y}{x}$v = y/x, $y^2=2x^2\ln |x|+A{x}^2$y² = 2x² ln|x| + Ax².

The usual slips

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

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

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