Identifying separable differential equations

[Probability 0.99]

I am unsure about whether the differential equation above is separable or not. At first, I believed it was and tried solving the differential equation using the separation method, it gave me the wrong answer. I could only obtain the right answer using the integrating factor method.

Why is this equation not separable? It appears X can be on one side and the other is just the function of t...

I don't major in Maths, so I am not sure where I have gone wrong when deciding whether this equation is separable or not.

Thanks

 

[MarkFL]

Yes, the equation is separable...we may write it as:

[MATH]\d{x}{t}=3x+18[/MATH]
or:

[MATH]\frac{1}{x+6}\,dx=3\,dt[/MATH]
Integrate:

[MATH]\int\frac{1}{x+6}\,dx=3\int\,dt[/MATH]
[MATH]\ln|x+6|=3t+C[/MATH]
This implies:

[MATH]x(t)=c_1e^{3t}-6[/MATH]

 

[Probability 0.99]

Yes, the equation is separable...we may write it as:

[MATH]\d{x}{t}=3x+18[/MATH]
or:

[MATH]\frac{1}{x+6}\,dx=3\,dt[/MATH]
Integrate:

[MATH]\int\frac{1}{x+6}\,dx=3\int\,dt[/MATH]
[MATH]\ln|x+6|=3t+C[/MATH]
This implies:

[MATH]x(t)=c_1e^{3t}-6[/MATH]

Thanks. I have noticed the mistake I have made now. Why did you change C to C1 at the end?

 

[MarkFL]

Thanks. I have noticed the mistake I have made now. Why did you change C to C1 at the end?

I mainly changed the parameter to show that it had been algebraically manipulated/changed. If we pick up here:

[MATH]\ln|x+6|=3t+C[/MATH]
This of course directly implies:

[MATH]x+6=\pm e^{3t+C}=\pm e^Ce^{3t}[/MATH]
Now, if we let:

[MATH]c_1=\pm e^C[/MATH]
And consider that we lost the trivial solution:

[MATH]x(t)\equiv-6[/MATH]
when we separated the variables, this means \(c_1\in\mathbb{R}\), and we may state:

[MATH]x+6=c_1e^{3t}[/MATH]
Hence:

[MATH]x(t)=c_1e^{3t}-6[/MATH]
Treating the ODE as a linear equation, we avoid all this. Later, you will learn a method call undetermined coefficients that will allow you to solve linear equations even more efficiently. You will be able to see by inspection that the homogeneous solution is:

[MATH]x_h(t)=c_1e^{3t}[/MATH]
And the particular solution must take the form:

[MATH]x_p(t)=A\implies x_p'(t)=0[/MATH] (here \(A\) is a constant)

And by substitution into the ODE we find:

[MATH]0-3A=18\implies A=-6[/MATH]
Thus:

[MATH]x_p(t)=-6[/MATH]
And, by the principle of superposition, we have the general solution:

[MATH]x(t)=x_h(t)+x_p(t)=c_1e^{3t}-6\quad\checkmark[/MATH]

 

[Probability 0.99]

I mainly changed the parameter to show that it had been algebraically manipulated/changed. If we pick up here:

[MATH]\ln|x+6|=3t+C[/MATH]
This of course directly implies:

[MATH]x+6=\pm e^{3t+C}=\pm e^Ce^{3t}[/MATH]
Now, if we let:

[MATH]c_1=\pm e^C[/MATH]
And consider that we lost the trivial solution:

[MATH]x(t)\equiv-6[/MATH]
when we separated the variables, this means \(c_1\in\mathbb{R}\), and we may state:

[MATH]x+6=c_1e^{3t}[/MATH]
Hence:

[MATH]x(t)=c_1e^{3t}-6[/MATH]
Treating the ODE as a linear equation, we avoid all this. Later, you will learn a method call undetermined coefficients that will allow you to solve linear equations even more efficiently. You will be able to see by inspection that the homogeneous solution is:

[MATH]x_h(t)=c_1e^{3t}[/MATH]
And the particular solution must take the form:

[MATH]x_p(t)=A\implies x_p'(t)=0[/MATH] (here \(A\) is a constant)

And by substitution into the ODE we find:

[MATH]0-3A=18\implies A=-6[/MATH]
Thus:

[MATH]x_p(t)=-6[/MATH]
And, by the principle of superposition, we have the general solution:

[MATH]x(t)=x_h(t)+x_p(t)=c_1e^{3t}-6\quad\checkmark[/MATH]

what does c1∈R mean? I major in Economics, sorry about my lack of knowledge. Thanks

 

[MarkFL]

what does c1∈R mean? I major in Economics, sorry about my lack of knowledge. Thanks

It's just shorthand to say that the parameter \(c_1\) can be any real number (is an element of the reals \(\mathbb{R}\)), including zero.