# Thread: Linear Problem - Slope of Curve Through Point

1. ## Linear Problem - Slope of Curve Through Point

Hi all,
I have started on differential equations today and need some help (as usual).

From what i have read so far.

The solution to a differential equation will be another function or differential equation.

For example the slope of a curve is a differential equation. The solution of this differential equation will actually give you the curve of the function or the equation of the function itself.

Now i have this problem to solve.

"find the curve whose slope at any point (x,y) is equal to 5y and which passes through the point (1,-2)".

Now the slope is given as 5y;

$\displaystyle slope=\frac{dy}{dx}=5y$

Now this is a differential equation in itself and the solution to the differential equation will give me the curve of the function itself.

rearranging the slope term

multiply across by dx:

$\displaystyle \frac{dy}{dx}.dx=5y.dx$

$\displaystyle dy=5y.dx$

To solve the differential equation i need to integrate both sides;

$\displaystyle \int dy=\int 5y.dx$

$\displaystyle x+C1= \frac{5y^2}{2} +C2$

rearranging similar terms:

$\displaystyle x- \frac{5y^2}{2} =C2-C1$

$\displaystyle x- \frac{5y^2}{2} =C.........equation (1)$

Now we are given the point (1,-2) so substitute into equation (1) for x and y

$\displaystyle (1)- \frac{5(-2)^2}{2} =C.........equation (1)$

$\displaystyle C=-9$

Now the solution to the problem is C inserted into equation (1)

$\displaystyle x- \frac{5y^2}{2} =-9$

or

$\displaystyle x- \frac{5y^2}{2} +9=0$

Now to check this if i substitute the solution back into the original equation for y and if i differentiate y with respect to x i should get the slope of 5y?

I think i understand it-some of the text overcomplicates this if you ask me(or more probably i have not got the brains to understand it),
can anybody give me a more basic explanation if i have not got it correct,
Is my method/solution correct?
Thanks
John

2. ## re: Linear Problem - Slope of Curve Through Point

Originally Posted by celtic1234
Hi all,
I have started on differential equations today and need some help (as usual).

From what i have read so far.

The solution to a differential equation will be another function or differential equation.

For example the slope of a curve is a differential equation. The solution of this differential equation will actually give you the curve of the function or the equation of the function itself.

Now i have this problem to solve.

"find the curve whose slope at any point (x,y) is equal to 5y and which passes through the point (1,-2)".

Now the slope is given as 5y;

$\displaystyle slope=\frac{dy}{dx}=5y$

Now this is a differential equation in itself and the solution to the differential equation will give me the curve of the function itself.

rearranging the slope term

multiply across by dx:

$\displaystyle \frac{dy}{dx}.dx=5y.dx$

$\displaystyle dy=5y.dx$

To solve the differential equation i need to integrate both sides;

$\displaystyle \int dy=\int 5y.dx$

$\displaystyle x+C1= \frac{5y^2}{2} +C2$
NO!! $\displaystyle \int 5y dy= 5y^2/2+ C$ but you cannot integrate y with respect to x! y is an unknown function of x.

Instead, you need to separate the x and y terms:
$\displaystyle \frac{dy}{y}= 5 dx$

Now integrate
$\displaystyle \int \frac{dy}{y}= \int 5 dx$

rearranging similar terms:

$\displaystyle x- \frac{5y^2}{2} =C2-C1$

$\displaystyle x- \frac{5y^2}{2} =C.........equation (1)$

Now we are given the point (1,-2) so substitute into equation (1) for x and y

$\displaystyle (1)- \frac{5(-2)^2}{2} =C.........equation (1)$

$\displaystyle C=-9$

Now the solution to the problem is C inserted into equation (1)

$\displaystyle x- \frac{5y^2}{2} =-9$

or

$\displaystyle x- \frac{5y^2}{2} +9=0$

Now to check this if i substitute the solution back into the original equation for y and if i differentiate y with respect to x i should get the slope of 5y?

I think i understand it-some of the text overcomplicates this if you ask me(or more probably i have not got the brains to understand it),
can anybody give me a more basic explanation if i have not got it correct,
Is my method/solution correct?
Thanks
John
No, your method is not correct. In particular, you cannot integrate y, an unknown function of x, with respect to x. $\displaystyle \int y dx$ is NOT the same as $\displaystyle \int y dy$.

3. ## Re: Linear Problem - Slope of Curve Through Point

ok,
so if i am getting what you re saying;

$\displaystyle \frac{1}{y}.dy=5.dx$

integrate both sides:

$\displaystyle \int \frac{1}{dy}.dy=\int 5.dx$

$\displaystyle ln y + C1= 5x + C2$

$\displaystyle ln y-5x=C$

At point (1,-2)

$\displaystyle -ln (-2)-5(1)=C$

$\displaystyle C =-5.69315$

Substituting into the solution for C

$\displaystyle ln y-5x=-5.69315$

$\displaystyle ln y-5x+5.69315 =0$

Is this it?
John