Question

By inspection, find a one parameter family of solutions of the differential equation xy' = 4y. Verify that each member of its family also satisfies the initial condition y(0) = 0

Answer 1

x y' -4y=0

y' - 4/x y=0

now we can better inspect

Answer 2

Ok thanks i get it so far, but I'm confused by the wording.. one parameter family.. i'm not sure where to go from there

Answer 3

since it say inspection, I don't think we have to try ti solve it

Answer 4

When they say by inspection, they mean just try to see through the problem to the solution without any serious work. It can be done. Note:
\[ x y' - 4y = 0 \rightarrow xy' = 4y\]
So taking the derivative of the function and multiplying by x is the same as multiplying the function by 4. To me, this suggests something of the form
\[ y = A x^4\]
where a is an arbitrary constant.

Answer 5

A*

Answer 6

OOo i think i get it now thank you!

Answer 7

The full solution, in contrast, is the following:
\[ xy' - 4y = 0\]
\[ y' - \frac{4}{x} y = 0\]
Introduce an integrating factor
\[e^{\mu(x)} \]
so
\[[ye^{\mu(x)} ]' = y'e^\mu + \mu'ye^\mu\]
Identifying
\[ \mu'e^\mu = -\frac{4}{x}e^\mu \rightarrow \mu = e^{-4 \ln(x)} = e^{\ln(x^{-4})} = x^{-4} \]
So multiplying through and grouping together,
\[ [x^{-4}y]' = 0\]
so
\[ x^{-4}y = A\]
and finally
\[y =Ax^4\]
That would be showing all the rigorous work... what we did is called by inspection, or just looking at it ;)

Answer 8

Alright that explains it, thank you!