Identifying separable differential equations

Probability 0.99

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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
 
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MarkFL

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

\(\displaystyle \d{x}{t}=3x+18\)

or:

\(\displaystyle \frac{1}{x+6}\,dx=3\,dt\)

Integrate:

\(\displaystyle \int\frac{1}{x+6}\,dx=3\int\,dt\)

\(\displaystyle \ln|x+6|=3t+C\)

This implies:

\(\displaystyle x(t)=c_1e^{3t}-6\)
 

Probability 0.99

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

\(\displaystyle \d{x}{t}=3x+18\)

or:

\(\displaystyle \frac{1}{x+6}\,dx=3\,dt\)

Integrate:

\(\displaystyle \int\frac{1}{x+6}\,dx=3\int\,dt\)

\(\displaystyle \ln|x+6|=3t+C\)

This implies:

\(\displaystyle x(t)=c_1e^{3t}-6\)
Thanks. I have noticed the mistake I have made now. Why did you change C to C1 at the end?
 

MarkFL

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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:

\(\displaystyle \ln|x+6|=3t+C\)

This of course directly implies:

\(\displaystyle x+6=\pm e^{3t+C}=\pm e^Ce^{3t}\)

Now, if we let:

\(\displaystyle c_1=\pm e^C\)

And consider that we lost the trivial solution:

\(\displaystyle x(t)\equiv-6\)

when we separated the variables, this means \(c_1\in\mathbb{R}\), and we may state:

\(\displaystyle x+6=c_1e^{3t}\)

Hence:

\(\displaystyle x(t)=c_1e^{3t}-6\)

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:

\(\displaystyle x_h(t)=c_1e^{3t}\)

And the particular solution must take the form:

\(\displaystyle x_p(t)=A\implies x_p'(t)=0\) (here \(A\) is a constant)

And by substitution into the ODE we find:

\(\displaystyle 0-3A=18\implies A=-6\)

Thus:

\(\displaystyle x_p(t)=-6\)

And, by the principle of superposition, we have the general solution:

\(\displaystyle x(t)=x_h(t)+x_p(t)=c_1e^{3t}-6\quad\checkmark\)
 

Probability 0.99

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

\(\displaystyle \ln|x+6|=3t+C\)

This of course directly implies:

\(\displaystyle x+6=\pm e^{3t+C}=\pm e^Ce^{3t}\)

Now, if we let:

\(\displaystyle c_1=\pm e^C\)

And consider that we lost the trivial solution:

\(\displaystyle x(t)\equiv-6\)

when we separated the variables, this means \(c_1\in\mathbb{R}\), and we may state:

\(\displaystyle x+6=c_1e^{3t}\)

Hence:

\(\displaystyle x(t)=c_1e^{3t}-6\)

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:

\(\displaystyle x_h(t)=c_1e^{3t}\)

And the particular solution must take the form:

\(\displaystyle x_p(t)=A\implies x_p'(t)=0\) (here \(A\) is a constant)

And by substitution into the ODE we find:

\(\displaystyle 0-3A=18\implies A=-6\)

Thus:

\(\displaystyle x_p(t)=-6\)

And, by the principle of superposition, we have the general solution:

\(\displaystyle x(t)=x_h(t)+x_p(t)=c_1e^{3t}-6\quad\checkmark\)
what does c1∈R mean? I major in Economics, sorry about my lack of knowledge. Thanks
 

MarkFL

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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.
 
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