Question

 What is the solution of the differential equation \[\sqrt {a + x} \dfrac{{dy}}{{dx}} + x = 0\]?
A. \[3y + 2\sqrt {a + x} \cdot \left( {x - 2a} \right) = 3c\]
B. \[3y + 2\sqrt {a + x} \cdot \left( {x + 2a} \right) = 3c\]
C. \[3y + \sqrt {a + x} \cdot \left( {x - 2a} \right) = 3c\]
D. None of these

Answer 1

Hint:
The given differential equation is the combination of 2 variables. Thus we will separate the variables and apply the integration formula to get the solution.

Formula used
Power rule of integration:
\[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]




Complete step-by-step answer:
Given differential equation is \[\sqrt {a + x} \dfrac{{dy}}{{dx}} + x = 0\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{x}{{\sqrt {a + x} }}\]
\[ \Rightarrow dy = - \dfrac{x}{{\sqrt {a + x} }}dx\]
Now add and subtract \[a\] from the numerator of the right side expression:
\[ \Rightarrow dy = - \dfrac{{x + a - a}}{{\sqrt {a + x} }} dx\]
Simplify the right-side expression:
\[ \Rightarrow dy = - \dfrac{{x + a}}{{\sqrt {a + x} }} dx + \dfrac{a}{{\sqrt {a + x} }} dx\]
\[ \Rightarrow dy = - \sqrt {a + x} dx + \dfrac{a}{{\sqrt {a + x} }} dx\]
Now taking integration on both sides:
\[ \Rightarrow \int {dy} = - \int {\sqrt {a + x} dx} + \int {\dfrac{a}{{\sqrt {a + x} }} dx} \] …..(i)

Assume that \[a + x = z\]
\[\Rightarrow dx = dz\]

Substitute \[a + x = z\] and \[dx = dz\] in equation (i)
\[ \Rightarrow \int {dy} = - \int {\sqrt z dz} + \int {\dfrac{a}{{\sqrt z }} dz} \]
Applying power rule:
\[ \Rightarrow y = - \dfrac{{{z^{1 + \dfrac{1}{2}}}}}{{\dfrac{3}{2}}} + a\dfrac{{{z^{1 - \dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + c\]
\[ \Rightarrow y = - \dfrac{{{z^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} + a\dfrac{{\sqrt z }}{{\dfrac{1}{2}}} + c\]
Simplify the above equation:
\[ \Rightarrow y = - \dfrac{{2{z^{\dfrac{3}{2}}}}}{3} + 2a\sqrt z + c\]
Substitute \[z = a + x\] in the above equation:
\[ \Rightarrow y = - \dfrac{{2{{\left( {a + x} \right)}^{\dfrac{3}{2}}}}}{3} + 2a\sqrt {a + x} + c\]
Multiply 3 both sides of the equation:
\[ \Rightarrow 3y = - 2{\left( {a + x} \right)^{\dfrac{3}{2}}} + 6a\sqrt {a + x} + 3c\]
Rewrite above equation:
\[ \Rightarrow 3y = - 2\left( {a + x} \right)\sqrt {a + x} + 6a\sqrt {a + x} + 3c\]
\[ \Rightarrow 3y = - 2\sqrt {a + x} \left( {a + x - 3a} \right) + 3c\]
\[ \Rightarrow 3y = - 2\sqrt {a + x} \left( {x - 2a} \right) + 3c\]
\[ \Rightarrow 3y + 2\sqrt {a + x} \left( {x - 2a} \right) = 3c\]
Hence option A is the correct option.

Note: Student often do mistake to integrate \[\dfrac{1}{{\sqrt z }}\]. They used by mistake the integration formula of \[\dfrac{1}{x}\]. They got \[y = - \dfrac{{{z^{1 + \dfrac{1}{2}}}}}{{\dfrac{3}{2}}} + a\log z + c\] which is incorrect. The correct answer is \[y = - \dfrac{{{z^{1 + \dfrac{1}{2}}}}}{{\dfrac{3}{2}}} + a\dfrac{{{z^{1 - \dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + c\].