[SOLVED]separable variables

shorty

New member
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! Markov

Member
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

shorty

New member
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
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

shorty

New member
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
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! 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

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• shorty

Prove It

Well-known member
MHB Math Helper
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...

• shorty