How do you find the particular solution to #yy'-e^x=0# that satisfies y(0)=4?

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Answer 1 · 2017-01-13T18:17:59

#y(x) = sqrt(2e^x+15)#

The differential equation:

#y*(dy)/(dx) -e^x =0 #

is separable, so we can solve it by separating he variables and integrating:

#y*(dy)/(dx) = e^x #

#ydy =e^xdx#

#int ydy = int e^xdx#

#y^2/2 = e^x+C#

#y^2 = 2e^x+C#

for #x=0# we have:

#y(0)^2 = e^0+C#

#16= 1+ C => C =15#

So the particular solution we are seeking is:

#y(x) = sqrt(2e^x+15)#