Find Sol for ODE dy/dx=(x+y+2)^2

[1s1]

Homework Statement

Find the general solution:
$$\frac{dy}{dx}=(x+y+3)^{2}$$

Homework Equations

The Attempt at a Solution

Methods I have learned: separation of variables, integrating factor for linear equations, exact equations, and substitution. I don't even know where to begin on this one. It has a $y^{2}$ term so it isn't linear; it isn't an exact equation; so there must be a substution. But it doesn't look anything like the substitution problems I've done. I expanded it, but that just seems to make it really long. Any hints or help would be greatly appreciated!

 

[fzero]

Obviously the difficulty is that we have $x+y$ appearing in the power on the RHS, so an obvious choice of substitution would involve this combination. Try $u=x+y+3$. Since this is linear, $dy/dx$ and $du/dx$ are simply related.

 

[1s1]

Thanks! Clearly, I don't understand how substitution works ... yet.

The only substitution we've learned so far is for Bernoulli's Equation, $\frac{dy}{dx}+P(x)y=f(x)y^{n}$

Where the substitution is $u=y^{1-n}$

I'll run $u=x+y+3$ through and I should get a linear equation like you said.

Sorry for the newbie ODE question.

 

[fzero]

Thanks! Clearly, I don't understand how substitution works ... yet.

The only substitution we've learned so far is for Bernoulli's Equation, $\frac{dy}{dx}+P(x)y=f(x)y^{n}$

Where the substitution is $u=y^{1-n}$

I'll run $u=x+y+3$ through and I should get a linear equation like you said.

Sorry for the newbie ODE question.

Very often substitution involves trial and error, using experience to suggest promising approaches.

After the substitution the equation is still not linear (it involves $u^$), but it is separable, which allows us to easily solve it.

 

[1s1]

Making the suggested substitution: $u=x+y+3$ and using:
$$\int \frac{1}{a^{2}+u^{2}}du=\frac{1}{a}tan^{-1}\frac{u}{a}+C$$
$$x=tan^{-1}(x+y+3)+C$$

The choice of substitution in this case seems similar to what you would choose when doing integration by substitution, so hopefully that will be a trend and will make picking the stubstution easier. I suppose practicing with different types of substitutions also develops experience on likely beneficial substitutions. It also seems appropriate to choose $u$ so that $x$ and $y$ are simply related. Thanks again fzero!