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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
 2 years ago
 2 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
 2 years ago
 2 years ago

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imranmeah91Best ResponseYou've already chosen the best response.0
x y' 4y=0 y'  4/x y=0 now we can better inspect
 2 years ago

SBest ResponseYou've already chosen the best response.1
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
 2 years ago

imranmeah91Best ResponseYou've already chosen the best response.0
since it say inspection, I don't think we have to try ti solve it
 2 years ago

Jemurray3Best ResponseYou've already chosen the best response.1
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.
 2 years ago

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

Jemurray3Best ResponseYou've already chosen the best response.1
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 ;)
 2 years ago

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