## initial-value problem

i have no idea on how to start this problem.

Find a solution of the initial-value problem

dy/dx= -y^2 , y(0)=.0625

please if someone can help me start it or explain it to me

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 Recognitions: Gold Member Homework Help Science Advisor Do you know what a separable differential equation is?
 you need to first use separation of variables, to get $$y(x)$$. So: $$-\frac{dy}{y^{2}} = dx$$ $$\int -\frac{dy}{y^{2}} = \int dx$$

## initial-value problem

no we just started covering DE on friday

 Recognitions: Gold Member Homework Help Science Advisor Well they're the type of differential equation that are the easiest to solve. If you've not yet seen how to solve this type of DE, why not just postpone the resolution of the problem til you've learned about it?
 in class all we went over was exponential growth and decay. but i dont remember going over initial-value problems.
 Recognitions: Gold Member Homework Help Science Advisor All an "initial value problem" is is a DE together with an "initial condition", i.e. a value of y(0). So just solve the differential equation, and set y(0)=0.0625. this will specify a concrete value of the intgegration constant. So, start with solving the differential equation like courtrigrad showed you. Can you do that?
 im not sure exactly how to solve the DE. What does dy and dx become? do they just simply become x and y?
 Recognitions: Gold Member Homework Help Science Advisor You don't know how to solve an integral?!
 i can solve integrals, but im confused about what the integral of dx becomes. i can solve the integral of -1/y^2 which is 1/y. i dont know what goes on with the integral of dy or dx.
 Recognitions: Gold Member Homework Help Science Advisor $\int dx$ is a notation for $\int 1dx$. Surely you can solve $\int 1dx$? It's only a matter of finding a function F(x) such that F'(x) = 1 and adding an "integration constant" C to it.
 oh so the integral of dx is simply x+c. but what is going on with the dy?
 Recognitions: Gold Member Homework Help Science Advisor The dy is only there to say that you must integrate with respect to y. $$\int -\frac{dy}{y^{2}}$$ is a notation for $$\int -\frac{1}{y^{2}} \ \ dy$$