Proof equality of an equation with exponentials.

[yungman]

Proof if A,B and C are non zero constant:
$$Ae^{jax}+Be^{jbx}=Ce^{jcx}\;\Rightarrow\; a=b=c$$
The answer from the book involve differentiating it twice and manipulate a, b and c to proof.

My question is if I differentiate it once:
$$\Rightarrow\;jaAe^{jax}+jbBe^{jbx}=jcCe^{jcx}$$
So if
$$Ae^{jax}+Be^{jbx}=Ce^{jcx}\;\hbox { and }\;jaAe^{jax}+jbBe^{jbx}=jcCe^{jcx}$$
Does that already proof a=b=c?

 

[Dick]

Proof if A,B and C are non zero constant:
$$Ae^{jax}+Be^{jbx}=Ce^{jcx}\;\Rightarrow\; a=b=c$$
The answer from the book involve differentiating it twice and manipulate a, b and c to proof.

My question is if I differentiate it once:
$$\Rightarrow\;jaAe^{jax}+jbBe^{jbx}=jcCe^{jcx}$$
So if
$$Ae^{jax}+Be^{jbx}=Ce^{jcx}\;\hbox { and }\;jaAe^{jax}+jbBe^{jbx}=jcCe^{jcx}$$
Does that already proof a=b=c?

It's not a proof until you say why you think that proves a=b=c.

 

[yungman]

It's not a proof until you say why you think that proves a=b=c.

Can I say if
$$Ae^{jax}+Be^{jbx}=Ce^{jcx}\;\hbox { and }\;jd(Ae^{jax}+jbBe^{jbx})=jcCe^{jcx}\Rightarrow;d=c$$
and if
$$\;jaAe^{jax}+jbBe^{jbx}=jd(Ae^{jax}+jbBe^{jbx})\Rightarrow\; a=b=d$$

Therefore a=b=c

Thanks
Alan

 

[Dick]

Can I say if
$$Ae^{jax}+Be^{jbx}=Ce^{jcx}\;\hbox { and }\;jd(Ae^{jax}+jbBe^{jbx})=jcCe^{jcx}\Rightarrow;d=c$$
and if
$$\;jaAe^{jax}+jbBe^{jbx}=jd(Ae^{jax}+jbBe^{jbx})\Rightarrow\; a=b=d$$

Therefore a=b=c

Thanks
Alan

I really don't know where those implications are coming from and I don't see how you got a 'd' out of an expression containing a, b and c.

 

[milesyoung]

You have written some conditional statements in the form of $P \Rightarrow Q$, but you haven't included any proof of whether they're true or not.

By direct proof, for instance, you have to show that P being true forces Q to be true.