y' - y = 1 (differential equations, initial value, y (0) = 0 )?

Question

Find the solution to the initial value problem??

i can never seem to understand how to do this when there is just a constant on the other side

plz help thanks

Answer 1

dy/dx - y =1



--> dy/dx = 1+y



--> dy/(1+y) = dx



integrate --> ln(1+y) = x + constant



y(0)=0 --> constant =0



so the solution should be ln(1+y) = x



or y =exp(x) - 1

Answer 2

first thing you need to do is get y' by itself. Do that by adding y to both sides, at which point you will have y' = 1 + y. Convert to Leibniz notation, and it becomes dy/dx = 1 + y.



multiply both sides by dx, so you have dy = (1 + y)dx.

divide by (1 + y), so you have dy/(1 + y) = dx and now you can just integrate both sides.



now you have ln(1 + y) = x + C, and then you can solve the initial value problem from there.

Answer 3

dy/dx = y + 1



dx = 1 / (y+1) dy



integral [ 1 dx ] = integral [ 1 / (y+1) dy ]



x = ln( y + 1) + C



Now for the initial value:



0 = ln(0 + 1) + C = ln(1) + C = 0 + C



C = 0

Answer 4

y = e^x - 1



you cannot use seperation of variables, you must use an intergrating factor.



This site has a very good worked example which is simple



http://www.ucl.ac.uk/mathematics/geomath/level2/deqn/de8.html



enjoy!

Answer 5

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