Question

Find the value of a in the differential equation \[\sin \dfrac{{dy}}{{dx}} = a\]with \[y\left( 0 \right) = 1\] from the given options.
A \[{\sin ^{ - 1}}\dfrac{{\left( {y - 1} \right)}}{x} = a\]
B \[\sin \dfrac{{\left( {y - 1} \right)}}{x} = a\]
C \[\sin \dfrac{{\left( {1 - y} \right)}}{{\left( {1 + x} \right)}} = a\]
D \[\sin \dfrac{y}{{\left( {1 + x} \right)}} = a\]

Answer 1

Hint: To find value of a, first separate the variables x and y. Then perform integration to find general solution. Then use initial value to find value of constant. Then find value of a using particular solution.

Formula Used: \[\int {1dx} = x + c\]
Where, c is an arbitrary constant.

Complete step by step solution: The given differential equation is \[\sin \dfrac{{dy}}{{dx}} = a\] with initial value \[y\left( 0 \right) = 1\].
First simplify the equation by multiplying \[{\sin ^{ - 1}}\]on both sides of the equation.
\[\dfrac{{dy}}{{dx}} = {\sin ^{ - 1}}a\]
Now, separate the variables x and y by multiplying \[dx\]on both sides of the equation.
\[dy = \left( {{{\sin }^{ - 1}}a} \right)dx\]
Integrate both sides of the equation.
\[\begin{array}{l}\int {dy} = \int {\left( {{{\sin }^{ - 1}}a} \right)dx} \\y = x{\sin ^{ - 1}}a + c\end{array}\]
So, the general solution of the equation is \[y = x{\sin ^{ - 1}}a + c\].
Use initial value \[y\left( 0 \right) = 1\] to find value of c. Substitute \[y = 1\]and \[x = 0\].
 \[\begin{array}{l}y = x{\sin ^{ - 1}}a + c\\1 = \left( 0 \right){\sin ^{ - 1}}a + c\\1 = c\end{array}\]
Substitute \[c = 1\]in general solution to find particular solution.
 \[y = x{\sin ^{ - 1}}a + 1\]
Isolate the a to find the value of a. subtract 1 from both sides of the equation.
\[\begin{array}{l}y = x{\sin ^{ - 1}}a + 1\\y - 1 = x{\sin ^{ - 1}}a\end{array}\]
Divide both sides of the equation by x.
\[\dfrac{{y - 1}}{x} = {\sin ^{ - 1}}a\]
Apply sine function on both sides of the equation.
\[\sin \dfrac{{y - 1}}{x} = a\]
So, the value of a is \[\sin \dfrac{{y - 1}}{x}\].

Option ‘B’ is correct

Note: Most often mistake happens in this type of question is that, considering a as variable. Also initial value can be substitute in general solution not in differential equation.