Solving a first order differential equation with initial conditions

[arhzz]

Homework Statement

Solve the equation

Relevant Equations

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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!

 

[Mark44]

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

You forgot to multiply the last term on the right side by e.

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!

I get the same result as Wolfram.

 

[arhzz]

You forgot to multiply the last term on the right side by e.

I get the same result as Wolfram.

Yea I get it too now,my bad.Thanks for pointing the error out.Cheers