# Differential Equation Substitution

• January 28th 2011, 06:46 AM
StaryNight
Differential Equation Substitution
I'm trying to solve the problem (x+y+1)^2 dy/dx + (x+y+1)^2 + x^3 = 0. I've tried the substituting for x+y+1, x+y and (x+y+1)^2 but all to no avail. Could somebody please give a suggestion? This problem apparently just requires a change of variables and separation.
• January 28th 2011, 06:51 AM
TheEmptySet
Quote:

Originally Posted by StaryNight
I'm trying to solve the problem (x+y+1)^2 dy/dx + (x+y+1)^2 + x^3 = 0. I've tried the substituting for x+y+1, x+y and (x+y+1)^2 but all to no avail. Could somebody please give a suggestion? This problem apparently just requires a change of variables and separation.

This equation is exact Note that

$\underbrace{(x+y+1)^2}_{M}dy+\underbrace{[(x+y+1)^2+x^3]}_{N}dx=0$

$M_x=2(x+y+1)=N_y$

Can you finish from here?
• January 28th 2011, 07:02 AM
StaryNight
Quote:

Originally Posted by TheEmptySet
This equation is exact Note that

$\underbrace{(x+y+1)^2}_{M}dy+\underbrace{[(x+y+1)^2+x^3]}_{N}dx=0$

$M_x=2(x+y+1)=N_y$

Can you finish from here?

I'm afraid i'm still confused...
• January 28th 2011, 07:14 AM
TheEmptySet
Quote:

Originally Posted by StaryNight
I'm afraid i'm still confused...

Exact differential equation - Wikipedia, the free encyclopedia
This should also be in your book.

An equation is exact if there is a function $f(x,y)$ such that

$\nabla f \cdot d\vec{r}=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}=0$

Where $\vec{r}=x\vec{i}+y\vec{j}$ is the position vector.

Since the mixed partials match we know such a function exists so

$\frac{\partial f}{\partial y}=(x+y+1)^2 \implies f(x,y)=\frac{1}{3}(x+y+1)^3+g(x)$

But we also know what the partial derivative with respect to $x$ must be so we can solve for $g(x)$

$\frac{\partial f}{\partial x}=(x+y+1)^2+g'(x)=(x+y+1)^2+x^3 \implies g(x)=\frac{1}{4}x^4$

This gives $\frac{1}{3}(x+y+1)^3+\frac{1}{4}x^4=c$
is an implicit solution to the equation.
• January 28th 2011, 07:33 AM
StaryNight
Quote:

Originally Posted by TheEmptySet
Exact differential equation - Wikipedia, the free encyclopedia
This should also be in your book.

An equation is exact if there is a function $f(x,y)$ such that

$\nabla f \cdot d\vec{r}=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}=0$

Where $\vec{r}=x\vec{i}+y\vec{j}$ is the position vector.

Since the mixed partials match we know such a function exists so

$\frac{\partial f}{\partial y}=(x+y+1)^2 \implies f(x,y)=\frac{1}{3}(x+y+1)^3+g(x)$

But we also know what the partial derivative with respect to $x$ must be so we can solve for $g(x)$

$\frac{\partial f}{\partial x}=(x+y+1)^2+g'(x)=(x+y+1)^2+x^3 \implies g(x)=\frac{1}{4}x^4$

This gives $\frac{1}{3}(x+y+1)^3+\frac{1}{4}x^4=c$
is an implicit solution to the equation.

So what exactly is the substituion you are meant to use?
• January 28th 2011, 07:50 AM
Ackbeet
You said in your OP that this problem "apparently just requires a change of variables and separation." Does that mean you are required to solve it that way? Or could you solve it using the exact differential method that TheEmptySet has outlined for you? Because the exact differential method that TheEmptySet has outlined for you is not a substitution method, nor is it a separation of variables idea.
• January 28th 2011, 08:02 AM
StaryNight
Quote:

Originally Posted by Ackbeet
You said in your OP that this problem "apparently just requires a change of variables and separation." Does that mean you are required to solve it that way? Or could you solve it using the exact differential method that TheEmptySet has outlined for you? Because the exact differential method that TheEmptySet has outlined for you is not a substitution method, nor is it a separation of variables idea.

The question reads "Solve by a change of variables and separation". I'm in my first year of university and haven't covered the method TheEmptySet proposed.
• January 28th 2011, 08:05 AM
Ackbeet
Hmm. Why don't you show your work for the substitution u = x + y + 1?
• January 28th 2011, 08:08 AM
Krizalid
yes, the same, it's pretty straightforward, and it yields a separable equation.
• January 28th 2011, 08:17 AM
StaryNight
Quote:

Originally Posted by Ackbeet
Hmm. Why don't you show your work for the substitution u = x + y + 1?

Having had another go at the problem with this substitution I've managed to get a solution.