Differential Equations

Given,  $y=Ax:$

So,  $y^{|}=\frac{Adx}{dx}=A$

Putting value of $y^{|}$ in L.H.S. of the given D.E.

L.H.S= $x{y}^{|}=xA=Ax=y$ =R.H.S

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

Given, y= √1 + x²

Given,  $f{x}^n$ is $y=cosx+c$

So,  $y^{|}=-sinx$

Putting the value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}+sinx=-sinx+sinx=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

Given,  $f{x}^n$ is $y=x^2+2x+c$

So,  $y^{|}=2x+2$

Substituting value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}-2x-2=2x+2-2x-2=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

Given $f{x}^n$ is $y=e^x+1$

Differentiating with $x$ we get,

$y^{|}=\frac{dy}{dx}=e^x$

Again,

$y^{\left|\right|}=\frac{d^{2}y}{d{x}^2}=e^x$

Substituting value of $y^{\left|\right|}$ and $y^{|}$ in the given D.E. we get

$L.H.S.=y^{\left|\right|}-y^{|}=e^x-e^x=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

The highest order derivative present in the given D.E. is $\frac{d^{2}y}{d{x}^2}$ and its order is 2.

$\therefore$ Option (A) is correct.

In the given D.E,

$sin\frac{dy}{dx}$ is a trigonometric function of derivative $\frac{dy}{dx}$ . So it is not a polynomial equation so its derivative is not defined.

Hence, Degree of the given D.E. is not defined.

$\therefore$ Option (D) is correct.

The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

As the given D.E. is polynomial equation in its derivative, its degree is 1.

The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

The given order derivative present in the D.E. is $y^{|}$ so its order is 1.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.