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2.5 Solutions by Substitutions: Class Notes

2.5a

We will use substitutions to solve DEs (and IVPs) of the following types.

$$(y^2+xy)dx+x^{2}dy=0$$

An equation of the form $M(x,y)dx+N(x,y)dy=0$ is a homogeneous equation, if the coefficient functions possess the property $f(tx,ty)=t^{\alpha }f(x,y)$ (with the same $\alpha$ for both functions).

Tip: Here $dx$ and $dy$ are differentials.

For the equation above, check the coefficient of $dx$:

$$\begin{array}{cc}\hfill {\displaystyle M(x,y)}& {\displaystyle =y^2+xy}\hfill \\ \hfill {\displaystyle M(tx,ty)}& {\displaystyle =(ty{)}^2+(tx)(ty)=t^2{y}^2+t^{2}xy}\hfill \\ \hfill {\displaystyle }& {\displaystyle =t^2(y^2+xy)=t^{2}M(x,y)}\hfill \end{array}$$

So $M$ has the property with $\alpha =2$. The same is true of $N(x,y)=x^2$, since $N(tx,ty)=t^2{x}^2=t^{2}N(x,y)$.

Substitution:

Use either

$$\begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =ux}\hfill \\ \hfill {\displaystyle dy}& {\displaystyle =xdu+udx}\hfill \end{array}\text{or}\begin{array}{cc}\hfill {\displaystyle x}& {\displaystyle =vy}\hfill \\ \hfill {\displaystyle dx}& {\displaystyle =ydv+vdy}\hfill \end{array}$$

$${\displaystyle \frac{dy}{dx}-y=e^x{y}^2}$$

The DE ${\displaystyle \frac{dy}{dx}+P(x)y=f(x)y^n}$ is called Bernoulli’s equation.

Tip: Here ${\displaystyle \frac{dy}{dx}}$ is the derivative, and $n\ne 1$ (when $n=1$ or $n=0$, the equation is already linear).

Substitution:

$$u=y^{1-n}$$

In the equation above, $n=2$, so

$${\displaystyle u=y^{-1}\Rightarrow y=u^{-1}\Rightarrow \frac{dy}{dx}=-\frac{1}{u^2}\frac{du}{dx}.}$$

The equation ${\displaystyle \frac{dy}{dx}=f(Ax+By+C)}$ can be converted to a separable equation using the substitution $u=Ax+By+C$.

Differentiate the substitution, then solve for the derivative ${\displaystyle \frac{dy}{dx}}$:

$${\displaystyle u=Ax+By+C\Rightarrow \frac{du}{dx}=A+B\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{B}(\frac{du}{dx}-A)}$$

$${\displaystyle \frac{dy}{dx}=\frac{3x+2y}{3x+2y+2}}$$

For this equation, let $u=3x+2y+2$, so ${\displaystyle \frac{du}{dx}=3+2\frac{dy}{dx}}$.

2.5b

Example

Solve the given differential equation by using an appropriate substitution.

$$(y^2+xy)dx+x^{2}dy=0$$

Solution

This is a homogeneous DE. We can use either $y=ux$ or $x=vy$; here we use $y=ux$, so $dy=xdu+udx$.

Substitute for $y$ and $dy$:

$$(u^2{x}^2+x·ux)dx+x^2(xdu+udx)=0$$

$$u^2{x}^{2}dx+u{x}^{2}dx+x^{3}du+x^{2}udx=0$$

Combine the $dx$ terms, and divide out $x^2$:

$$(u^2+2u)dx+xdu=0\text{(separable)}$$

Separate the variables:

$${\displaystyle \frac{du}{u^2+2u}=-\frac{1}{x}dx}$$

Use partial fractions (PFD) on the left side: ${\displaystyle \frac{1}{u(u+2)}=\frac{1/2}{u}-\frac{1/2}{u+2}}$.

$${\displaystyle \int (\frac{1/2}{u}-\frac{1/2}{u+2})du=\int (-\frac{1}{x})dx}$$

$${\displaystyle \frac{1}{2}\ln \left|\frac{u}{u+2}\right|=-\ln |x|+\ln c_1=\ln \left|\frac{c_1}{x}\right|}$$

Multiply by 2 and exponentiate (the constant $c_1^2$ is renamed $c$):

$${\displaystyle \frac{u}{u+2}=\frac{c}{x^2}}$$

Back-substitute: $y=ux$, so ${\displaystyle u=\frac{y}{x}}$.

$${\displaystyle \frac{y/x}{y/x+2}=\frac{c}{x^2}\Rightarrow \frac{y}{y+2x}=\frac{c}{x^2}}$$

(Multiply the top and bottom of the left side by $x$.)

2.5c

Example

$${\displaystyle \frac{dy}{dx}-y=e^x{y}^2}$$

Solution

This is a Bernoulli equation with $n=2$.

$$u=y^{1-n}\Rightarrow u=y^{-1}\Rightarrow y=u^{-1}$$

$${\displaystyle \frac{dy}{dx}=-\frac{1}{u^2}\frac{du}{dx}}$$

Substitute for ${\displaystyle \frac{dy}{dx}}$, $y$, and $y^2$:

$${\displaystyle -\frac{1}{u^2}\frac{du}{dx}-\frac{1}{u}=e^x\frac{1}{u^2}}$$

Multiply by $-u^2$. The result is linear:

$${\displaystyle \frac{du}{dx}+u=-e^x}$$

$P(x)=1$, so the integrating factor is ${\displaystyle \mu =e^{\int dx}=e^x}$.

$${\displaystyle \frac{d}{dx}\left(e^{x}u\right)=-e^{2x}}$$

$${\displaystyle e^{x}u=-\frac{1}{2}{e}^{2x}+c_1}$$

$${\displaystyle u=-\frac{1}{2}{e}^x+c_1{e}^{-x}=-\frac{e^x}{2}+\frac{c_1}{e^x}=\frac{-e^{2x}+c}{2e^x}}$$

Here $c=2c_1$.

Back-substitute ${\displaystyle u=\frac{1}{y}}$:

$${\displaystyle \frac{1}{y}=\frac{c-e^{2x}}{2e^x}}$$

Example

$${\displaystyle 3(1+t^2)\frac{dy}{dt}=2ty(y^3-1)}$$

Solution

We want the form ${\displaystyle \frac{dy}{dt}+P(t)y=f(t)y^n}$. Distribute, and move the $y$ term to the left:

$${\displaystyle 3(1+t^2)\frac{dy}{dt}+2ty=2t{y}^4}$$

$${\displaystyle \frac{dy}{dt}+\frac{2t}{3(1+t^2)}y=\frac{2t}{3(1+t^2)}{y}^4}$$

This is a Bernoulli equation with $n=4$.

$$u=y^{1-4}=y^{-3}\Rightarrow y=u^{-1/3}$$

Then

$${\displaystyle \frac{dy}{dt}=-\frac{1}{3}{u}^{-4/3}\frac{du}{dt},y^4=u^{-4/3}.}$$

Substitute:

$${\displaystyle -\frac{1}{3}{u}^{-4/3}\frac{du}{dt}+\frac{2t}{3(1+t^2)}{u}^{-1/3}=\frac{2t}{3(1+t^2)}{u}^{-4/3}}$$

Multiply by $-3u^{4/3}$ (this changes the signs of the other two terms). The result is linear:

$${\displaystyle \frac{du}{dt}-\frac{2t}{1+t^2}u=-\frac{2t}{1+t^2}}$$

$${\displaystyle \mu =e^{\int -\frac{2t}{1+t^2}dt}=e^{-\ln (1+t^2)}=\frac{1}{1+t^2}}$$

$${\displaystyle \frac{d}{dt}\left[\frac{1}{1+t^2}u\right]=-\frac{2t}{{(1+t^2)}^2}}$$

$${\displaystyle \frac{1}{1+t^2}u=\frac{1}{1+t^2}+c}$$

$$u=1+c(1+t^2)$$

Back-substitute $u=y^{-3}$:

This implicit form is fine. Explicitly,

$$y={[1+c(1+t^2)]}^{-1/3}.$$

2.5d

Example

Solve the given initial-value problem.

$${\displaystyle \frac{dy}{dx}=\frac{3x+2y}{3x+2y+2},y(-1)=-1}$$

Solution

Let $u=3x+2y+2$. Then

$${\displaystyle \frac{du}{dx}=3+2\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{2}(\frac{du}{dx}-3)}$$

Since $3x+2y=u-2$, the DE becomes

$${\displaystyle \frac{1}{2}(\frac{du}{dx}-3)=\frac{u-2}{u}=1-\frac{2}{u}}$$

$${\displaystyle \frac{du}{dx}=\frac{2u-4}{u}+3=\frac{5u-4}{u}}$$

Separate the variables:

$${\displaystyle \int \frac{u}{5u-4}du=\int dx}$$

Long division gives ${\displaystyle \frac{u}{5u-4}=\frac{1}{5}+\frac{4/5}{5u-4}}$:

$${\displaystyle \int (\frac{1}{5}+\frac{4/5}{5u-4})du=\int dx}$$

(For the second term, substitute $t=5u-4$, $dt=5du$.)

$${\displaystyle \frac{1}{5}u+\frac{4}{25}\ln |5u-4|=x+c_1}$$

Multiply by 5:

$${\displaystyle u+\frac{4}{5}\ln |5u-4|=5x+c_2}$$

Back-substitute $u=3x+2y+2$ (so $5u-4=15x+10y+6$):

$${\displaystyle 3x+2y+2+\frac{4}{5}\ln |15x+10y+6|=5x+c_2}$$

Move $3x$ and the 2 to the right side and combine the constants:

$${\displaystyle 2y-2x+\frac{4}{5}\ln |15x+10y+6|=c}$$

Apply $y(-1)=-1$: when $x=-1$, $y=-1$, so ${\displaystyle c=0+\frac{4}{5}\ln |-19|=\frac{4}{5}\ln 19}$.