## S 3 years ago 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

1. imranmeah91

x y' -4y=0 y' - 4/x y=0 now we can better inspect

2. S

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

3. imranmeah91

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

4. Jemurray3

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.

5. Jemurray3

A*

6. S

OOo i think i get it now thank you!

7. Jemurray3

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 ;)

8. S

Alright that explains it, thank you!