Finding the equation of a curve

[Deimantas]

Homework Statement

The tangent of any point that belongs to a curve, cuts Y axis in such a way, that the cut off segment in Y axis is twice as big as the X value of the point. Find the equation of the curve, if point (1,4) is part of it.

Homework Equations

The ultimate solution is $y = x(4-lnx^2)$

The Attempt at a Solution

This might be a little confusing, so I'll try to clarify the situation: assuming for example, that point (1,4) is part of the curve, if we try to find the tangent of the curve at that point, we know that it will cut Y axis where y=2x=2. So in that case the tangent is a line going through (1,4) and (0,2).I guess we get $y = x(4-lnx^2)$ when we put x=1 and y=4 into the general solution, which gives us the specific constant value, then insert the constant value into the general solution to get the final solution. But I don't know how to obtain the general solution. Could you please help me solve this problem?

 

[Simon Bridge]

The curve is y=f(x)
In general: the slope of the tangent at point x=q is f'(q)
The equation of the line is y=f'(q)x+c(q) so that c(q) is the y intercept.
The problem statement is saying that c(q)=2q... which, in general, will be a function of q.

 

[Deimantas]

The curve is y=f(x)
In general: the slope of the tangent at point x=q is f'(q)
The equation of the line is y=f'(q)x+c(q) so that c(q) is the y intercept.
The problem statement is saying that c(q)=2q... which, in general, will be a function of q.

So we have $y=y'x+2x$ which is the same as $y'-y/x=-2$ which is a linear differential equation, solving it by using the Bernoulli method y=uv, we get the general solution $y=x(C-lnx^2)$. Inserting x=1, y=4 gives C=4, thus the final answer is $y=x(4-lnx^2)$.
You're awesome, thanks!