Trig Substitution

• Jan 11th 2007, 04:52 PM
turtle
Trig Substitution
I don't know how to do this problem, so if someone could please help me, I'll be grateful:

Solve the initial value problem

(x2 + 1) squared dy/dx = the square root of(x2 + 1) , y(0) = 1

Thanks
• Jan 11th 2007, 05:27 PM
ThePerfectHacker
Quote:

Originally Posted by turtle
I don't know how to do this problem, so if someone could please help me, I'll be grateful:

Solve the initial value problem

(x2 + 1) squared dy/dx = the square root of(x2 + 1) , y(0) = 1

Thanks

$\displaystyle (x^2+1)y'=\sqrt{x^2+1}$

$\displaystyle y'=\frac{\sqrt{x^2+1}}{x^2+1}=\frac{1}{\sqrt{1+x^2 }}$

$\displaystyle y=\int \frac{1}{\sqrt{1+x^2}}dx=\sinh^{-1}(x)+C$
$\displaystyle y(0)=1$
$\displaystyle 1=\sinh^{-1}(0)+C$
$\displaystyle 1=0+C$
$\displaystyle C=1$

$\displaystyle y=\sinh^{-1}(x)+1$
• Jan 11th 2007, 08:44 PM
Soroban
Hello, turtle!

Your title was "Trig Substitution" . . .

Quote:

Solve the initial value problem:

$\displaystyle (x^2 + 1)^2\,\frac{dy}{dx} \:= \:\sqrt{x^2 + 1}\qquad y(0) = 1$

We have: .$\displaystyle \frac{dy}{dx} \:=\:\frac{\sqrt{x^2+1}}{(x^2+1)^2}\quad\Rightarro w\quad dy \:=\:\frac{dx}{(x^2+1)^{\frac{3}{2}}}$

Integrate: .$\displaystyle \int dy \;=\;\int\frac{dx}{(x^2+1)^{\frac{3}{2}}}$

Let $\displaystyle x = \tan\theta\quad\Rightarrow\quad dx = \sec^2\theta\,d\theta$
. . $\displaystyle (x^2 + 1)^{\frac{3}{2}} \:=\:\left(\tan^2\theta + 1\right)^{\frac{3}{2}} \:=\:\left(\sec^2\theta\right)^{\frac{3}{2}} \:=\:\sec^3\theta$

Substitute: .$\displaystyle y(x) \;=\;\int\frac{\sec^2\theta\,d\theta}{\sec^3\theta } \;= \;\int\frac{d\theta}{\sec\theta} \;=\;\int\cos\theta\,d\theta \;=\;\sin\theta + C$

Back-substitute: .$\displaystyle \tan\theta = x\quad\Rightarrow\quad\sin\theta = \frac{x}{\sqrt{x^2+1}}$

. . $\displaystyle y(x) \;= \;\frac{x}{\sqrt{x^2+1}} + C$

Since $\displaystyle y(0) = 1\!:\;\;\frac{0}{\sqrt{0^2 + 1}} + C \:=\:1\quad\Rightarrow\quad C = 1$

Therefore: .$\displaystyle y(x) \;=\;\frac{x}{\sqrt{x^2+1}} + 1$