- #1

- 248

- 49

- Homework Statement
- Solve the equation

- Relevant Equations
- -

Hello!

Consider this ODE;

$$ x' = sin(t) (x+2) $$ with initial conditions x(0) = 1;

Now I've solved it and according to wolfram alpha it is correct (I got the homogenous and the particular solution)

$$ x = c * e^{-cos(t)} -2 $$ and now I wanted to plug in the initial conditions and this is how i did it;

$$ 1 = c * e^{-cos(0)} -2 $$ now we can rewrite that as ## 1 = \frac{c}{e} -2 ## and now we want c we multiply by e

$$ e = c -2 $$ c should be c = e +2

Now I plug that back into my x

$$ x = e + 2 * e^{-cos(t)} -2 $$ and I get ## x = e * e^{-cos(t)} ##

Now wolfram alpha is giving me a diffrent result; ## x = 3e^{-cos(t)+1} -2 ##

I don't see how they get to this? I am not sure if I am doing something wrong,could it be that wolfram alpha is not really solving the ODE I input?

Thanks!

Consider this ODE;

$$ x' = sin(t) (x+2) $$ with initial conditions x(0) = 1;

Now I've solved it and according to wolfram alpha it is correct (I got the homogenous and the particular solution)

$$ x = c * e^{-cos(t)} -2 $$ and now I wanted to plug in the initial conditions and this is how i did it;

$$ 1 = c * e^{-cos(0)} -2 $$ now we can rewrite that as ## 1 = \frac{c}{e} -2 ## and now we want c we multiply by e

$$ e = c -2 $$ c should be c = e +2

Now I plug that back into my x

$$ x = e + 2 * e^{-cos(t)} -2 $$ and I get ## x = e * e^{-cos(t)} ##

Now wolfram alpha is giving me a diffrent result; ## x = 3e^{-cos(t)+1} -2 ##

I don't see how they get to this? I am not sure if I am doing something wrong,could it be that wolfram alpha is not really solving the ODE I input?

Thanks!