Welcome to our community

Be a part of something great, join today!

[SOLVED] separable variables

shorty

New member
Feb 5, 2012
16
I have a question that is stumping me. I'd be grateful on some assistance.

Show that the substitutions $z= ax + by + c$ changes $y' = f(ax + by + c)$ into an equation with separable variables. Hence, solve the equation $y' = (x+y)^2$.

(hint: $\int \frac{1}{(1 + u^2)}du = tan^{-1} u+c$)

I thought i could do this, but my working takes me nowhere near to the hint. Therefore i'm lost. Help plz!:confused:
 

Markov

Member
Feb 1, 2012
149
If $t=x+y\implies \dfrac{dt}{dx}=1+\dfrac{dy}{dx}$ so the ODE becomes $\dfrac{dt}{dx}-1=t^2,$ which is separable.
 

shorty

New member
Feb 5, 2012
16
Thanks,

But how does that tie in to the 1st part which says to show the substitutions z=ax + by + c changes y' = f(ax + by + c) into an equation with separable variables...??


If $t=x+y\implies \dfrac{dt}{dx}=1+\dfrac{dy}{dx}$ so the ODE becomes $\dfrac{dt}{dx}-1=t^2,$ which is separable.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Thanks,

But how does that tie in to the 1st part which says to show the substitutions z=ax + by + c changes y' = f(ax + by + c) into an equation with separable variables...??
Can you see that $ \displaystyle x + y $ is of the form $ \displaystyle ax + by + c $?
 

shorty

New member
Feb 5, 2012
16
yes. but how am i using the z substitution to show anything..?

Can you see that $ \displaystyle x + y $ is of the form $ \displaystyle ax + by + c $?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I have a question that is stumping me. I'd be grateful on some assistance.

Show that the substitutions $z= ax + by + c$ changes $y' = f(ax + by + c)$ into an equation with separable variables. Hence, solve the equation $y' = (x+y)^2$.

(hint: $\int \frac{1}{(1 + u^2)}du = tan^{-1} u+c$)

I thought i could do this, but my working takes me nowhere near to the hint. Therefore i'm lost. Help plz!:confused:
put \(z=ax+by+c\) then:

\[\large \frac{dz}{dx}=a+by' \]

so:

\[ \large z'-a=bf(z)\]

which is what Markov was trying to point you in the direction of.

CB
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
yes. but how am i using the z substitution to show anything..?
So if you let $ \displaystyle z = x + y $, then

\[ \displaystyle \begin{align*} y &= z - x \\ \frac{dy}{dx} &= \frac{dz}{dx} - 1 \end{align*} \]

So substitute this into the DE...

\[ \displaystyle \begin{align*} \frac{dy}{dx} &= (x + y)^2 \\ \frac{dz}{dx} - 1 &= z^2 \\ \frac{dz}{dx} &= 1 + z^2 \\ \frac{1}{1 + z^2}\,\frac{dz}{dx} &= 1 \end{align*}\]

So clearly the equation is separable.

This is the exact same method Markov showed you, just using z instead of t...