- #1
twiztidmxcn
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This question is an initial value problem for diffeq. We are asked to solve explicitly for y.
(1+cos(x))dy = ((e^(-y))+1)*sin(x)dx , y(0) = 0
I attempted a separation of variables and ended up with the following:
dy / ((e^(-y))+1) = (sin(x) / (1 + cos(x))) dx
I know that my next step is to integrate both sides and then solve using the given initial value, but I am unsure as to how I am supposed to integrate either side.
For the right side, I believe I can integrate using u-substitution, where:
u = cos(x) + 1
du / dx = -sin(x)
So that the right side becomes -1/u, integrates to -ln(u), then -ln(1+cos(x)).
For the left side, I've tried using partial fraction decomposition but end up either with my original equation or the natural logarithm of a negative number.
This is where I need help, is in the integration of the left hand side.
thank you
-twiztidmxcn
(1+cos(x))dy = ((e^(-y))+1)*sin(x)dx , y(0) = 0
I attempted a separation of variables and ended up with the following:
dy / ((e^(-y))+1) = (sin(x) / (1 + cos(x))) dx
I know that my next step is to integrate both sides and then solve using the given initial value, but I am unsure as to how I am supposed to integrate either side.
For the right side, I believe I can integrate using u-substitution, where:
u = cos(x) + 1
du / dx = -sin(x)
So that the right side becomes -1/u, integrates to -ln(u), then -ln(1+cos(x)).
For the left side, I've tried using partial fraction decomposition but end up either with my original equation or the natural logarithm of a negative number.
This is where I need help, is in the integration of the left hand side.
thank you
-twiztidmxcn