Question

Find the equation of a curve passing through the point (0, 2), given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer 1

Given:

The sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

To find:

Find the equation of a curve passing through the point (0, 2), given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Solution:

According to the given information, let F(x,y) be the curve and dy/dx be the slope to the curve.

Hence dy/dx+5=x+y

Let us rewrite the equation as dy/dx−y=x+5

The equation is of the form dy/dx+Py=Q

Here P=−1 and Q=x+5

∫−dx=−x  

Hence I.F=e^{−x}

The required solution is ye−x=∫(x+5).e^{−x}.dx+C

∫(x+5)e^{−x}dx can be done by parts.

Let u=x+5, hence du=dx and dv=e^{−x}dx;v=−e^{−x}

∫(x+5)e^{−x}=(x+5)(−e^{−x})−∫(−e^{−x})dx

=−(x+5)e^{−x}+∫e^{−x}dx

=−(x+5)e^{−x}−e^{−x}+C

Hence the required solution is ye−x=−(x+5)e^{−x}−e^{−x}+C

ye^{−x}=(5−x)e^{−x}−e^{−x}+C

ye^{−x}=(4−x)e^{−x}+C

To evaluate C, let us substitute the values of x and y in the above equation.

2e^0=(4−0)e^0+C

C=−2

Substituting for C we get,

ye^{−x}=(4−x)e^{−x}−2

dividing throughout by e^{−x} we get,

y=(4−x)−2/e^{−x}

y=x−4−2e^{x} is the required solution.