find the particular solution of the differential equation that satisfies the initial condition

Question

zubhanwc3 · Answer

$$\sqrt{x}+\sqrt{y}y'=0$$$$y(1) = 4$$

zubhanwc3 · Answer

@zepdrix

zepdrix · Answer

So this is separable, yes?

zepdrix · Answer

Subtract sqrt x from each side,
"multiply" the dx over,$$\Large\rm \sqrt y ~dy=-\sqrt x~ dx$$

zubhanwc3 · Answer

these were 2 equations that were in the same line for the problem

zepdrix · Answer

Understand how to integrate each side?

zubhanwc3 · Answer

i have a general idea, but could you go over it, im in online school, and my calc foundation is very weak :/

zepdrix · Answer

$$\Large\rm \int\limits\sqrt y ~dy=\int\limits -\sqrt x~ dx$$So you want to `rationalize` the terms before you integrate.$$\Large\rm \int\limits y^{1/2} ~dy=\int\limits -x^{1/2}~ dx$$Remember that from your rules of exponents I hope? :O
Then we just apply the Power Rule for Integration to each term.$$\Large\rm \color{royalblue}{\int\limits x^{n}~dx=\frac{1}{n+1}x^{n+1}}$$

zubhanwc3 · Answer

thank you, i know what to do from here.

zepdrix · Answer

cool c: