
If $y=Ce^{\sin^{-1}x}$, then corresponding to this differential equation is
A. $\dfrac{\text{d}y}{\text{d}x}=\dfrac{y}{\sqrt{1-x^{2}}}$
B. $\dfrac{\text{d}y}{\text{d}x}=\dfrac{1}{\sqrt{1-x^{2}}}$
C. $\dfrac{\text{d}y}{\text{d}x}=\dfrac{x}{\sqrt{1-x^{2}}}$
D. None of these.
Answer
164.7k+ views
Hint: Find the derivative of the given equation using the necessary formula wherever it is required. If you see the term that can be substituted using the base equation, then substitute it. This will give our required differential equation.
Formula Used: $\dfrac{\text{d}}{\text{d}x}\lgroup\sin^{-1}x\rgroup= \dfrac{1}{\sqrt{1-x^{2}}}$
Complete step by step solution: The given equation is,
$y=Ce^{\sin^{-1}x}$.........(i)
We know,
$\dfrac{\text{d}}{\text{d}x}\lgroup\sin^{-1}x\rgroup= \dfrac{1}{\sqrt{1-x^{2}}}$........(ii)
Now, differentiating (i) with respect to x, we get
$\dfrac{\text{d}y}{\text{d}x}$= $Ce^{\sin^{-1}x}$ $\times$ $1/$ $\sqrt{1-x^2}$ ....... $(iii) using (ii)$
Comparing equation (i) with equation (ii) and doing the necessary substitution, we get
$\Rightarrow\dfrac{\text{d}y}{\text{d}x}= \dfrac{y}{\sqrt{1-x^{2}}}$
The above equation is our required differential equation.
Option ‘A’ is correct
Additional Information: A differential equation in mathematics is an expression containing one or more derivatives of a function given by $dy/dx$. It is described as the equation including derivatives of one or more dependent variables with regard to one or more independent variables. The primary goal of differential equations is to analyze the solutions that satisfy the equations and the attributes of the answers.
Note: While differentiating make sure to write the differentiation value of $\sin^{-1}x$ (which students can get by formula) instead of writing the differentiated value of $\sin~x$ because many do the mistakes writing the differentiation of $\sin^{-1}x$ as $\cos~x$.
Formula Used: $\dfrac{\text{d}}{\text{d}x}\lgroup\sin^{-1}x\rgroup= \dfrac{1}{\sqrt{1-x^{2}}}$
Complete step by step solution: The given equation is,
$y=Ce^{\sin^{-1}x}$.........(i)
We know,
$\dfrac{\text{d}}{\text{d}x}\lgroup\sin^{-1}x\rgroup= \dfrac{1}{\sqrt{1-x^{2}}}$........(ii)
Now, differentiating (i) with respect to x, we get
$\dfrac{\text{d}y}{\text{d}x}$= $Ce^{\sin^{-1}x}$ $\times$ $1/$ $\sqrt{1-x^2}$ ....... $(iii) using (ii)$
Comparing equation (i) with equation (ii) and doing the necessary substitution, we get
$\Rightarrow\dfrac{\text{d}y}{\text{d}x}= \dfrac{y}{\sqrt{1-x^{2}}}$
The above equation is our required differential equation.
Option ‘A’ is correct
Additional Information: A differential equation in mathematics is an expression containing one or more derivatives of a function given by $dy/dx$. It is described as the equation including derivatives of one or more dependent variables with regard to one or more independent variables. The primary goal of differential equations is to analyze the solutions that satisfy the equations and the attributes of the answers.
Note: While differentiating make sure to write the differentiation value of $\sin^{-1}x$ (which students can get by formula) instead of writing the differentiated value of $\sin~x$ because many do the mistakes writing the differentiation of $\sin^{-1}x$ as $\cos~x$.
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