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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
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

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imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.0x y' 4y=0 y'  4/x y=0 now we can better inspect

S
 3 years ago
Best ResponseYou've already chosen the best response.1Ok 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

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.0since it say inspection, I don't think we have to try ti solve it

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1When 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.

S
 3 years ago
Best ResponseYou've already chosen the best response.1OOo i think i get it now thank you!

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1The 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 ;)

S
 3 years ago
Best ResponseYou've already chosen the best response.1Alright that explains it, thank you!
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