Solving differential equations using a change of variable

**Erghhh** · March 21st 2009, 07:41 AM

My book is telling me

d/du((e^u)dy/dx)

=(e^u)dy/dx + (e^u)(d2y/dx2).(dx/du)

I understand where the first part has come from, but not sure how the second part was derived. Would anyone kindly provide me with an explanation?

**ADARSH** · March 21st 2009, 08:43 AM

Originally Posted by Erghhh

My book is telling me

d/du((e^u)dy/dx)

=(e^u)dy/dx + (e^u)(d2y/dx2).(dx/du)

I understand where the first part has come from, but not sure how the second part was derived. Would anyone kindly provide me with an explanation?

This will make it clear

$\frac{d}{du}\{e^u\frac{dy}{dx}\}$

$= \frac{dy}{dx}\frac{d}{du}(e^u) +e^u \frac{d}{du}(\frac{dy}{dx})$

Differnetiation of dy/dx wrt u will be given by (using chain rule)

$= \frac{y'}{dx} \cdot \frac{du}{dx}$
------------------------------------------
So differnetiation will be

$=\frac{dy}{dx}e^u + e^u \frac{d^2(y)}{dx^2} \cdot \frac{dx}{du}$

Thread: Solving differential equations using a change of variable

LinkBack

Thread Tools

Display

Solving differential equations using a change of variable

Erghhhhhhh....:p

Similar Math Help Forum Discussions

Differential equations with one variable

Solving for a variable with quadratic equations

Differential equations help , rate of change

Solving differential equations using a change of variable

Solving limits by change of variable

Search Tags