Tag: Ordinary Differential Equations

Prove a method to obtain a first-order, separable differential equation

by RoRi

January 31, 2016

Prove that the substitution $y = \frac{x}{v}$ transforms a homogeneous equation $y' = f(x,y)$ into a first-order, separable equation for $v$ .

Proof. Let

$y = \frac{x}{v} \quad \implies \quad y' = \frac{v - xv'}{v^2}.$

Since

$f(x,y) = f \left(x, \frac{x}{v} \right)$

is homogeneous, we have

$f \left( xt, \frac{x}{v} t \right) = f\left( x, \frac{x}{v} \right).$

Letting $t = \frac{1}{x}$ we then have

$f(x,y) = f \left( 1, \frac{1}{v} \right).$

So,

$\begin{align*} y' = f(x,y) && \implies && \frac{v-xv'}{v^2} &= f \left( 1, \frac{1}{v} \right) \\ && \implies && v'x &= v - v^2 f \left( 1, \frac{1}{v} \right) \\ && \implies && \frac{1}{v - v^2 f \left( 1, \frac{1}{v} \right)} v' &= \frac{1}{x}. \end{align*}$

Hence, this equation is separable $. \qquad \blacksquare$

Find an implicit formula satisfied by solutions of y y′ = e^{x + 2y} sin x

by RoRi

January 31, 2016

Assume solutions of the equation

$yy' = e^{x+2y} \sin x$

exist and find an implicit formula satisfied by these solutions.

This is a first order separable equation. We compute

$\begin{align*} yy' = e^{x + 2y} \sin x && \implies && ye^{-2y} y' &= e^x \sin x \\ && \implies && \int y e^{-2y} \, dy &= \int e^x \sin x \, dx \\ && \implies && -\frac{y}{2} e^{-2y} - \frac{1}{4} e^{-2y} &= \frac{1}{2}\big(e^x \sin x - e^x \cos x\big) + C \\ && \implies && e^{-2y} \left( -y-\frac{1}{2} \right) &= e^x (\sin x - \cos x) + C \\ && \implies && e^{-2y} \left( y + \frac{1}{2} \right) &= e^x (\cos x - \sin x) + C. \end{align*}$

Both of the integrals were evaluated using integration by parts. Also, we already established a formula for $\int e^x \sin x \, dx$ in this exercise (Section 6.17, Exercise #20, taking $a = b = 1$ ).

Find an implicit formula satisfied by solutions of xyy′ = 1 + x² + y² + x²y²

by RoRi

January 31, 2016

Assume solutions of the equation

$xyy' = 1 + x^2 + y^2 + x^2 y^2$

exist and find an implicit formula satisfied by these solutions.

This is a first-order separable equation. We compute

$\begin{align*} xyy' = 1 + x^2 + y^2 + x^2 y^2 && \implies && y' \left( \frac{y}{1+y^2} \right) &= x + \frac{1}{x} \\ && \implies && \int \frac{y}{1+y^2} \, dy &= \int \left( x + \frac{1}{x} \right) \, dx \\ && \implies && \frac{1}{2} \log (1+y^2) &= \frac{1}{2} x^2 + \log |x| + C \\ && \implies && 1+y^2 &= Cx^2 e^{x^2}. \end{align*}$

Find an implicit formula satisfied by solutions of (x² – 4) y′ = y

by RoRi

January 31, 2016

Assume solutions of the equation

$(x^2 - 4)y' = y$

exist and find an implicit formula satisfied by these solutions.

This a first-order separable equation. We compute

$\begin{align*} (x^2 - 4)y' = y && \implies && \frac{y'}{y} &= \frac{1}{x^2-4} \\ && \implies && \int \frac{1}{y} \, dy &= \int \frac{1}{x^2-4} \, dx \\ && \implies && \log |y| &= \int \frac{1}{(x-2)(x+2)} \, dx \\ && \implies && \log |y| &= \frac{1}{4} \int \frac{1}{x-2} \, dx - \frac{1}{4} \int \frac{1}{x+2} \, dx \\ && \implies && \log |y| &= \frac{1}{4} \log |x-2| - \frac{1}{4} \log |x+2| + C \\ && \implies && 4 \log |y| &= \log |x-2| - \log|x+2| + C \\ && \implies && y^4 &= C \left( \frac{x-2}{x+2} \right) \\ && \implies && y^4(x+2) &= C(x-2). \end{align*}$

Find an implicit formula satisfied by solutions of xy(1 + x²) y′ – (1 + y²) = 0

by RoRi

January 31, 2016

Assume solutions of the equation

$xy (1 + x^2) y' - (1+y^2) = 0$

exist and find an implicit formula satisfied by these solutions.

This is a separable first-order equation. We compute

$\begin{align*} xy (1+x^2) y' - (1+y^2) = 0 && \implies && y' \left( \frac{y}{1+y^2} \right) &= \frac{1}{x(1+x^2)} \\ && \implies && \int \frac{y}{1+y^2} \, dy &= \int \frac{1}{x(1+x^2)} \, dx \\ && \implies && \frac{1}{2} \log (1+y^2) &= \int \frac{1}{x} \,dx - \int \frac{x}{1+x^2} \, dx \\ && \implies && \frac{1}{2} \log (1+y^2) &= \log |x| - \frac{1}{2} \log (1+x^2) + C \\ && \implies && (1+y^2)(1+x^2) &= Cx^2. \end{align*}$

Find an implicit formula satisfied by solutions of (1 – x²)^1/2 y′ + 1 + y² = 0

by RoRi

January 31, 2016

Assume solutions of the equation

$(1-x^2)^{\frac{1}{2}} y' + 1 + y^2 = 0$

exist and find an implicit formula satisfied by these solutions.

This is a separable first-order equation. We compute

$\begin{align*} (1-x^2)^{\frac{1}{2}} y' - (1+y^2) = 0 && \implies && \frac{-y'}{1+y^2} &= \frac{1}{\sqrt{1-x^2}} \\ && \implies && -\int \frac{1}{1+y^2} \, dy &= \int \frac{1}{\sqrt{1-x^2}} \, dx \\ && \implies && -\arctan y &= \arcsin x + C \\ && \implies && \arctan y + \arcsin x &= C. \end{align*}$

Find an implicit formula satisfied by solutions of (x – 1) y′ = xy

by RoRi

January 31, 2016

Assume solutions of the equation

$(x-1)y' = xy$

exist and find an implicit formula satisfied by these solutions.

This is a separable first-order equation. We compute

$\begin{align*} (x-1)y' = xy && \implies && \frac{y'}{y} &= \frac{x}{x-1} \\ && \implies && \int \frac{1}{y} \, dy &= \int \frac{x}{x-1} \, dx \\ && \implies && \log |y| &= -\log|x-1| + x + C \\ && \implies && y &= Ce^x (x-1). \end{align*}$

Find an implicit formula satisfied by solutions of y (1 – x²)^1/2 y′ = x

by RoRi

January 31, 2016

Assume solutions of the equation

$y \sqrt{1-x^2} y' = x$

exist and find an implicit formula satisfied by these solutions.

This is a separable first order equation. We compute

$\begin{align*} y \sqrt{1-x^2} y' = x && \implies && yy' &= \frac{x}{\sqrt{1-x^2}} \\[9pt] && \implies && \int y \, dy &= \int \frac{x}{\sqrt{1-x^2}} \, dx \\[9pt] && \implies && \frac{1}{2} y^2 &= -\sqrt{1-x^2} + C \\[9pt] && \implies && y^2 + 2 \sqrt{1-x^2} &= C. \end{align*}$

Find an implicit formula satisfied by solutions of y′ = (y-1)(y-2)

by RoRi

January 31, 2016

Assume solutions of the equation

$y' = (y-1)(y-2)$

exist and find an implicit formula satisfied by these solutions.

This is a separable first order equation. We compute

$\begin{align*} y' = (y-1)(y-2) && \implies && \frac{y'}{(y-1)(y-2)} &= 1 \\[9pt] && \implies && \int \frac{1}{(y-1)(y-2) \, dy} &= \int \, dx \\[9pt] && \implies && \log |y-2| - \log |y-1| &= x + C \\ && \implies && \frac{y-2}{y-1} &= Ce^x \\ && \implies && y-2 &= C(y-1)e^x. \end{align*}$

Find an implicit formula satisfied by solutions of (x + 1) y′ + y² = 0

by RoRi

January 31, 2016

Assume solutions of the equation

$(x+1)y' + y^2 = 0$

exist and find an implicit formula satisfied by these solutions.

This is a separable first order equation. We compute

$\begin{align*} (x+1)y' + y^2 = 0 && \implies && y' \left( \frac{1}{y^2} \right) &= \frac{-1}{x+1} \\[9pt] && \implies && \int \frac{1}{y^2} \, dy &= -\int \frac{1}{x+1} \, dx \\[9pt] && \implies && -\frac{1}{y} &= -\log |x+1| + C \\[9pt] && \implies && y \big( \log |x+1| + C \big) &= 1. \end{align*}$