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Solution by Substitution 1

https://isaacphysics.org/questions/solution_by_substitution1?stage=a_level

I managed to solve this question but I dont think I did it with the right method.

In the first part of my working out (before the dashed line) I feel pretty confident in the method, but after that to be honest I didnt really know what I was doing.

I just used some form of trial and error and it led me to the right answer, but I dont think this was the intended way.
If anyone could help that would be greatly appreciated.

Working:

Original post by mosaurlodon

Working:

Youve got the main idea in there, and its a case of noting the left hand side is the derivative of y^2. You could note that the derivative of the right hand side is
LHS + 1
so that suggests modifying the ODE by adding 1 to both sides so
2y dy/dx + 1 = y^2+x + 1
Then define
z = y^2 + x
so the ode becomes fairly simple and could be solved using seperation of variables or a simple linear first order ode or ...

Reply 3

DFranklin

Original post by mosaurlodon

In the first part of my working out (before the dashed line) I feel pretty confident in the method, but after that to be honest I didnt really know what I was doing.
I just used some form of trial and error and it led me to the right answer, but I dont think this was the intended way.
If anyone could help that would be greatly appreciated.

I would expect IsaacPhysics expects you to know how to solve a first order equation of the form

\dfrac{dy}{dx} + P(x) y = Q(x) \quad (\dagger)

https://en.wikipedia.org/wiki/Integrating_factor#Solving_first_order_linear_ordinary_differential_equations

In this particular case, you'd rewrite as dk/dx - k = x and then multiply by

e^-x

to get

\dfrac{dk}{dx}e^{-x} - ke^{-x} = xe^{-x}

and then note the LHS is

\dfrac{d}{dx}\left(ke^{-x}\right)

, so integrating both sides w.r.t. x gives the desired result.

mqb's solution is neater, but I think the **intent** was "make a substitution to get it in the form