Mutated Integrating Factors

Big Idea

To solve the linear ODE

$\displaystyle \frac{\mathrm dy}{\mathrm dx} + P(x)y = Q(x),$

we compute the integrating factor

$I(x) := e^{\int P(x)\, \mathrm dx},$

then obtain the general solution

$\displaystyle I(x) \cdot y = \int I(x) Q(x)\, \mathrm dx,$

where we need to evaluate the right-hand side. For a derivation of this solving technique, click here.

Questions

Question 1. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} - 2xy = xe^{x^2}.$

(Click for Solution)

Solution. We first identify $P(x) = -2x$ and $Q(x) = xe^{x^2}$ . By the method of integrating factors, compute

$I(x) = e^{\int P(x)\, \mathrm dx} = e^{\int -2x\, \mathrm dx} = e^{-x^2}.$

Then the general solution is given by

$\begin{aligned} I(x) y &= \int I(x)Q(x)\, \mathrm dx \\ e^{-x^2} y &= \int e^{-x^2} \cdot xe^{x^2}\, \mathrm dx \\ &= \int x\, \mathrm dx \\ &=\frac{x^2}{2} + C.\end{aligned}$

Question 2. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} - \frac yx = xe^x,\quad x > 0.$

(Click for Solution)

Solution. We first identify $P(x) = -1/x$ and $Q(x) = xe^x$ . By the method of integrating factors, compute

$I(x) = e^{\int P(x)\, \mathrm dx} = e^{\int -\frac 1x\, \mathrm dx} = \displaystyle e^{-{\ln |x|}} = \frac 1{|x|} = \frac 1x.$

Then the general solution is given by

$\begin{aligned} I(x) y &= \int I(x)Q(x)\, \mathrm dx \\ \frac 1x \cdot y &= \int \frac 1x \cdot xe^{x}\, \mathrm dx \\ &= \int e^x\, \mathrm dx \\ &= e^x + C.\end{aligned}$

Question 3. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} + y = y^3.$

Hint. Use the substitution $u = y^{-2}$ .

(Click for Solution)

Solution. By the given substitution, applying the chain rule yields

$\displaystyle \frac{\mathrm du}{\mathrm dx} = -2 y^{-3} \cdot \frac{\mathrm dy}{\mathrm dx}.$

Dividing the original equation by $y^3$ then multiplying by $-2$ ,

$\displaystyle -2y^{-3} \cdot \frac{\mathrm dy}{\mathrm dx} -2y^{-2} = -2.$

Substituting, we obtain

$\displaystyle \frac{\mathrm du}{\mathrm dx} - 2u = -2.$

We identify $P(x) = -2$ and $Q(x) = -2$ . By the method of integrating factors, compute

$I(x) = e^{\int P(x)\, \mathrm dx} = e^{\int -2\, \mathrm dx} = e^{-2x}.$

Then the general solution is given by

$\begin{aligned} I(x) \cdot u &= \int I(x)Q(x)\, \mathrm dx \\ e^{-2x} \cdot u &= \int e^{-2x} \cdot -2\, \mathrm dx \\ &=e^{-2x} + C.\end{aligned}$

Back-substituting $u = y^{-2}$ yields the general solution

$e^{-2x} \cdot y^{-2} = e^{-2x} + C.$

Exercise. Try to solve the simplified differential equation using the method of separable variables too.

This is an example of a Bernoulli ODE. Details here.

Question 4. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{y}{x+y^3},\quad y > 0.$

(Click for Solution)

Solution. By the inverse of a derivative,

$\begin{aligned} \frac{\mathrm dx}{\mathrm dy} &= \frac{x+y^3}{y} = \frac xy + y^2. \end{aligned}$

Hence,

$\displaystyle \frac{\mathrm dx}{\mathrm dy} - \frac 1y \cdot x = y^2.$

We identify $P(y) = -1/y$ and $Q(y) = y^2$ . By the method of integrating factors, compute

$I(y) = e^{\int P(y)\, \mathrm dx} = e^{\int - \frac 1y\, \mathrm dy} = e^{-{\ln|y|} } =\displaystyle \frac{1}{|y|} = \frac 1y.$

Then the general solution is given by

$\begin{aligned} I(y) \cdot x &= \int I(y)Q(y)\, \mathrm dy \\ \frac 1y \cdot x &= \int \frac 1y \cdot y^2\, \mathrm dy \\ &= \int y\, \mathrm dy \\ \frac{x}{y} &= \frac{y^2}{2} + C.\end{aligned}$

—Joel Kindiak, 18 Apr 25, 1138H

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