Now, let's recall the definition of degree of Differential equation.
Degree of differential equation :- The degree of differential equation is defined as the degree of highest order derivative occur in the given differential equation when all the other derivatives are free from radicals and fractions.
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Question :- What is the degree of the following differential equation ?
[tex]\rm \: x = \sqrt{1 + \dfrac{ {d}^{2}y}{ {dx}^{2} } } \\ \\ [/tex]
[tex] \: \: \: \: \: \: (a) \: \: 1 \\ [/tex]
[tex] \: \: \: \: \: \: (b) \: \: 2 \\ [/tex]
[tex] \: \: \: \: \: \: (c) \: \: 3 \\ [/tex]
[tex] \: \: \: \: \: \: (d)\rm \: \: \: degree \: not \: def \: ined \\ [/tex]
[tex] \\[/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given differential equation is
[tex]\rm \: x = \sqrt{1 + \dfrac{ {d}^{2}y}{ {dx}^{2} } } \\ \\ [/tex]
On squaring both sides, we get
[tex]\rm \: {x}^{2} =1 + \dfrac{ {d}^{2}y}{ {dx}^{2} } \\ \\ [/tex]
can be further rewritten as
[tex]\rm \: {x}^{2} =1 + \bigg(\dfrac{ {d}^{2}y}{ {dx}^{2} }\bigg)^{1} \\ \\ [/tex]
Now, let's recall the definition of degree of Differential equation.
Degree of differential equation :- The degree of differential equation is defined as the degree of highest order derivative occur in the given differential equation when all the other derivatives are free from radicals and fractions.
So, using this definition,
[tex]\bf \: Degree \:of \: differential \: equation \: = \: 1 \\ \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
Order of differential equation :- Order of differential equation is defined as highest order derivative occur in the given differential equation.
Question :-
97. What is the degree of the following differential equation?
[tex] \qquad\qquad \displaystyle\rm \[ x=\sqrt{1+\frac{d^{2} y}{d x^{2}}} \][/tex]
(a) 1
(b) 2
(c) 3
(d) Degree is not defined
Answer :-
Given:- [tex]\displaystyle \rm x=\sqrt{1+\frac{d^{2} y}{d x^{2}}}[/tex]
To find:- Degree of given equation.
Solution :-
[tex] \[ \begin{array}{l} \\ \\ \displaystyle\rm \therefore x=\sqrt{1+\frac{d^{2} y}{d x^{2}}} \\\\ \\ \\ \displaystyle\rm \Rightarrow x^{2}=1+ \left \lgroup\frac{d^{2} y}{d x^{2}} \right \rgroup ^ \red{ {1} } \qquad\qquad [ \because \: \: \text{ Squareing both sides. }] \end{array} \] \\ \\ \\ [/tex]
So, degree is 1 .
Hence, degree is 1.
Correct option is A.