First-Order Differential Equation - help

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Dinoduck94

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May 15, 2019
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In a exam test paper we have the below question:

Find an expression for 'y' in terms of 'x' if:

(x-1)y' = 3y

I have tried to solve this, but I think my understanding went off somewhere because when I put this into Symbolab to check my answer I get something different.

My solution was to start by moving all the 'y' terms to the left hand side:

1/y * (x-1)y' = 3

Then by moving the 'x' terms to the right:

1/y * y' = 3/(x-1)
Simplified:
y'/y = 3/(x-1)

When I integrate this, I get the below expression:

ln(y) = 3*ln(x-1)+c1

And then finally, to get the expression for 'y', I raise the right hand side to be the power of 'e':

y = e(3*ln(x-1)+c1)

However, Symbolab gives me:

y = ec1x3 - 3ec1x2 + 3ec1x - ec1

I haven't got a paid subscription with Symbolab, so I can't see the steps to achieving this.

Can anyone help me see where I went wrong, please?
 
Dinoduck94 said:
In a exam test paper we have the below question:

Find an expression for 'y' in terms of 'x' if:

(x-1)y' = 3y

I have tried to solve this, but I think my understanding went off somewhere because when I put this into Symbolab to check my answer I get something different.

My solution was to start by moving all the 'y' terms to the left hand side:

1/y * (x-1)y' = 3

Then by moving the 'x' terms to the right:

1/y * y' = 3/(x-1)
Simplified:
y'/y = 3/(x-1)

When I integrate this, I get the below expression:

ln(y) = 3*ln(x-1)+c1

And then finally, to get the expression for 'y', I raise the right hand side to be the power of 'e':

y = e(3*ln(x-1)+c1)

However, Symbolab gives me:

y = ec1x3 - 3ec1x2 + 3ec1x - ec1

I haven't got a paid subscription with Symbolab, so I can't see the steps to achieving this.

Can anyone help me see where I went wrong, please?
Click to expand...
You needed a bit more algebraic manipulation.

ln(y) = 3*ln(x-1)+c1

ln(y) =ln(x-1)3+ln(ec1)

ln(y) =ln[ec1*(x-1)3]

Now remove ln & expand (x-1)3 and Continue....
 
Subhotosh Khan said:
You needed a bit more algebraic manipulation.

ln(y) = 3*ln(x-1)+c1

ln(y) =ln(x-1)3+ln(ec1)

ln(y) =ln[ec1*(x-1)3]

Now remove ln & expand (x-1)3 and Continue....
Click to expand...

Thank you for your time.

Can you explain how 3*ln(x-1) + c1 turns into ln(x-1)3+ln(ec1) , please?

Thanks
 
Dinoduck94 said:
Thank you for your time.

Can you explain how 3*ln(x-1) + c1 turns into ln(x-1)3+ln(ec1) , please?

Thanks
Click to expand...
Are you reviewing this with paper/pencil or staring at the screen?

Using rules (& definition) of ln function:

ln(ax) = x * ln(a) ...... &

ln(e) = 1

To able to be proficient at DE - you will have to be master at these manipulations (algebra is pre-req for calculus for a reason).
 
Subhotosh Khan said:
Are you reviewing this with paper/pencil or staring at the screen?

Using rules (& definition) of ln function:

ln(ax) = x * ln(a) ...... &

ln(e) = 1

To able to be proficient at DE - you will have to be master at these manipulations (algebra is pre-req for calculus for a reason).
Click to expand...

That is a rule worth putting in my notes, thank you.

My apologies, I misunderstood initially what you wrote. I thought you were raising the natural logarithm of (x-1) to the power of 3:

ln(x-1)3 ,

But I see you meant the Natural Logarithm of (x-1)3

ln((x-1)3)

Thanks for your help
 
An alternate approach would be to write the ODE as:

[MATH]\frac{dy}{dx}-\frac{3}{x-1}y=0[/MATH]
Compute integrating factor:

[MATH]\mu(x)=\exp\left(-3\int \frac{1}{x-1}\,dx\right)=(x-1)^{-3}[/MATH]
And so the ODE becomes:

[MATH](x-1)^{-3}\frac{dy}{dx}-3(x-1)^{-4}y=0[/MATH]
[MATH]\frac{d}{dx}\left((x-1)^{-3}y\right)=0[/MATH]
[MATH](x-1)^{-3}y=c_1[/MATH]
Hence, the general solution is given by:

[MATH]y(x)=c_1(x-1)^3[/MATH]
 
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