Question

How do you solve the differential $ \dfrac{{dy}}{{dx}} = 4x + \dfrac{{4x}}{{\sqrt {16 - {x^2}} }} $ ?

Answer 1

Hint: To solve this problem we should know about the basic property of differential and how integration of a given function.
Integration: In simple terms it is calculating the area under the curve of given function.
Properties of integration:
  $ \int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} $

Complete step by step solution:
To solve it let's move one differential term to the right hand side so the whole equation changes into an integration term.
Moving $ dx $ to right hand side.
  $ dy = \left( {4x + \dfrac{{4x}}{{\sqrt {16 - {x^2}} }}} \right)dx $
Taking integration on both sides. We get,
  $ \Rightarrow \int {dy} = \int {\left( {4x + \dfrac{{4x}}{{\sqrt {16 - {x^2}} }}} \right)dx} $
  $ \Rightarrow \int {1.dy} = \int {\left( {4xdx + \dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} \right)} $
By doing integration as flowing general rule in the left hand side and separate right hands term and go for the same.
  $ \Rightarrow \int {1.dy} = \int {4xdx} + \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} $
  $ \Rightarrow y = \dfrac{{4{x^{1 + 1}}}}{{1 + 1}} + \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} = 2{x^2} + \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} + C $ …….. $ (1) $
By using, $ \int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} $
Taking $ \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} $ as separate and integrate it with respect to $ x $ .
  $ \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} = \int {\dfrac{{4x}}{{\sqrt u }}dx} $
Lets take $ 16 - {x^2} $ as $ u $ and differentiate with respect to $ x $ we get,
  $ \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {16 - {x^2}} \right) = - 2x $
  $ du = - 2xdx $
Hence, keeping back in above integration. We get,
  $ \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} = - \int {2{u^{ - \dfrac{1}{2}}}du} $
By doing its integration we get,
  $ \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} = - \int {2\dfrac{{{u^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}dx} = - 2\dfrac{{{u^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} = - 4{u^{\dfrac{1}{2}}} $
By keeping back value of $ u $ in it we get,
  $ \int {\dfrac{{4x}}{{\sqrt {16 - {x^2}} }}dx} = - 4{(16 - {x^2})^{\dfrac{1}{2}}} $
Keeping it back in eq $ (1) $ . we get,
In place of all constants we can keep a single constant.
  $ \Rightarrow y = 2{x^2} + - 4{(16 - {x^2})^{\dfrac{1}{2}}} + C $
Hence, the resultant equation will be $ y = 2{x^2} + - 4{(16 - {x^2})^{\dfrac{1}{2}}} + C $ .
So, the correct answer is “$ y = 2{x^2} + - 4{(16 - {x^2})^{\dfrac{1}{2}}} + C $”.

Note: The differential equation is used in daily life. The study of it is a wide field in pure and applied mathematics, physics, and engineering. All of these disciplines are concerned with the properties of differential equations of various types. In biology and economics, differential equations are used to model the behaviour of complex systems.