Solving differential equations
When integrating simple expressions, the constant of integration, the \(+ c\) term, may remain an unknown. The value of \(c\) can be worked out when additional information is given in the question, .
Example (extension)
Find the equation of the curve for which \(\frac{{dy}}{{dx}} = 4{x^3} + 6{x^2}\) and which passes through the point \((1,3)\).
Solution
\(y = \int {(4{x^3}} + 6{x^2})\,\,dx = {x^4} + 2{x^3} + c\)
Substituting \(x = 1\) and \(y = 3\) (from the coordinate point given in the question):
\(y = {x^4} + 2{x^3} + c\)
\(3 = {1^4} + 2{(1)^3} + c\)
\(3 = 3 + c\)
\(c = 0\)
Therefore the equation of the curve is \(y = {x^4} + 2{x^3}\)
Question
Extension
Find the equation of the curve for which \(\frac{{dy}}{{dx}} = 2x + 1\) and which passes through the point \((2,9)\).
\(y = \int {(2x + 1)} \,\,dx = {x^2} + x + c\)
Substituting \(x = 2\) and \(y = 9\):
\(9 = 4 + 2 + c\)
\(c = 3\)
\(y = {x^2} + x + 3\)
Therefore the equation of the curve is \(y = {x^2} + x + 3\)
Question
Extension
The gradientIn a graph, the gradient is the steepness of the line. The greater the gradient, the greater the rate of change. of a tangent to a curve is given as \(f\textquotesingle(x)= 6x - \frac{5}{{{x^2}}}\). Find the equation of the curve if it passes through the point \((1,6)\).
\(f(x) = \int {6x-\frac{5}{{{x^2}}}}\,\,dx\)
\(= \int {(6x}-5{x^{ - 2}})\,\,dx\)
\(f(x) = \frac{{6{x^2}}}{2} - \frac{{5{x^{ - 1}}}}{{ - 1}} + c\)
\(f(x) = 3{x^2} + \frac{5}{x} + c\)
Substitute when \(x = 1\) and \(y = 6\):
\(6 = 3{(1)^2} + \frac{5}{1} + c\)
\(6 = 8 + c\)
\(c = - 2\)
Therefore the equation of the curve is \(f(x) = 3{x^2} + \frac{5}{x} - 2\)
