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- Homework Statement
- Find two solutions for the initial condition y(0)=1.

- Relevant Equations
- Initial condition: y(0)=1

I have an ODE, which is y' = 5x

My attempt at solving this problem is as follows:

Separate the equation, we get, dy/(y-1)

Integrate both sides, ∫ dy/(y-1)

We are left with, (5/4)(y-1)

Now we may solve for C using our initial condition y(0) = 1. Substituting each of the values into (5/4)(y-1)

We can now isolate for y which gives us y = x

Therefore, the two solutions are y = x

Is this correct? Did I solve correctly? Thank you.

^{3}(y-1)^{1/5}with the initial condition y(0)=1, I must find two solutions.My attempt at solving this problem is as follows:

Separate the equation, we get, dy/(y-1)

^{1/5}= 5x^{3}dx.Integrate both sides, ∫ dy/(y-1)

^{1/5}= ∫ 5x^{3}dx.We are left with, (5/4)(y-1)

^{4/5}= (5x^{4})/4 + C.Now we may solve for C using our initial condition y(0) = 1. Substituting each of the values into (5/4)(y-1)

^{4/5}= (5x^{4})/4 + C gives us 0 = 0 + C.We can now isolate for y which gives us y = x

^{20/4}- 1.Therefore, the two solutions are y = x

^{20/4}- 1 and y = 1 or would it be y = 0?Is this correct? Did I solve correctly? Thank you.

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