Chapter 9 Differential Equations

\begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0 \end{align}

\begin{align} \Rightarrow y^{m\;'}+\;\sin(y^m)\;=0 \end{align}

The highest order derivative present in the differential equation is y^{m '}. Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

The given differential equation is:

y' + 5y = 0

The highest order derivative present in the differential equation isy'. Therefore, its order is one.

It is a polynomial equation in y'. The highest power raised to y' is 1. Hence, its degree is one.

\begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align}

The highest order derivative present in the given differential equation is\begin{align}\frac{d^2s}{dt^2}.\end{align}

 Therefore, its order is two. It is a polynomial equation in

\begin{align}\frac{d^2s}{dt^2} and \frac{ds}{dt}.\end{align}

The power raised to is 1.  \begin{align} \frac{d^2s}{dt^2} \end{align}

\begin{align}\left(\frac{d^2y}{dx^2}\right)^2\;+\;cos\left(\frac{dy}{dx}\right)\;=\;0\end{align}

The highest order derivative present in the given differential equation is \begin{align}\frac{d^2y}{dx^2}.\end{align}

Therefore, its order is 2. The given differential equation is not a polynomial equation in its derivatives.

Hence, its degree is not defined.

\begin{align}\frac{d^2y}{dx^2}=\cos3x  + sin3x\end{align}

\begin{align}\Rightarrow\frac{d^2y}{dx^2} - \cos3x  - sin3x = 0\end{align}

The highest order derivative present in the differential equation is\begin{align}\frac{d^2y}{dx^2}.\end{align}

Therefore, its order is two.It is a polynomial equation in \begin{align}\frac{d^2y}{dx^2}\end{align} 

and the power raised to is 1. 

\begin{align}\frac{d^2y}{dx^2}\end{align}

Hence, its degree is one.

(y^m)² + (yⁿ)³ + (y')⁴ + y⁵ =0

The highest order derivative present in the differential equation isy^m. Therefore, its order is three.

The given differential equation is a polynomial equation in y^{m ,}  y^{n , y'.}

The highest power raised to y^m is 2. Hence, its degree is 2.

The highest order derivative present in the differential equation is y^m. Therefore, its order is three.

It is a polynomial equation in y^m , yⁿ  and y' . The highest power raised to y^m is 1. Hence, its degree is 1.

y^' + y =e^x

y^' + y  - e^x =0

The highest order derivative present in the differential equation is y^'. Therefore, its order is one.

The given differential equation is a polynomial equation in y^' and the highest power raised to y^' is one. Hence, its degree is one.

yⁿ + (y')² + 2y =0

The highest order derivative present in the differential equation is yⁿ. Therefore, its order is two.

The given differential equation is a polynomial equation in yⁿ and y' and the highest power raised to yⁿ is one.

Hence, its degree is one.

yⁿ + 2y^' + siny = 0

The highest order derivative present in the differential equation is yⁿ. Therefore, its order is two.

This is a polynomial equation in yⁿ and y' and the highest power raised to yⁿ is one. Hence, its degree is one.

\begin{align}\left(\frac{d^2y}{dx^2}\right)^3\;+ \left(\frac{dy}{dx}\right)^2+\;sin\left(\frac{dy}{dx}\right)\;+ 1=\;0\end{align}

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

\begin{align}2x^2\frac{d^2y}{dx^2}\;- \;3\frac{dy}{dx}\;+ y=\;0\end{align}

The highest order derivative present in the given differential equation is

\begin{align}\frac{d^2y}{dx^2}\end{align}

Therefore, its order is two.

Hence, the correct answer is A.

y = e^x +1

Differentiating both sides of this equation with respect to x, we get:

\begin{align}\frac{dy}{dx}=\frac{d}{dx}(e^x + 1)\end{align}

=> y^' = e^x                          ...(1)

Now, differentiating equation (1) with respect to x, we get:

\begin{align}\frac{d}{dx}(y^{'})=\frac{d}{dx}(e^x)\end{align}

=> y^'' = e^x

Substituting the values of y^' and y^'' in the given differential equation, we get the L.H.S. as:

y^'' - y^' = e^x - e^x = 0 = R.H.S.

Thus, the given function is the solution of the corresponding differential equation.

y = x² + 2x + C

Differentiating both sides of this equation with respect to x, we get:

\begin{align}y^{'}=\frac{d}{dx}(x^2 + 2x + C)\end{align}

=> y^' = 2x + 2

Substituting the value of y^' in the given differential equation, we get:

L.H.S. = y^' - 2x - 2 = 2x + 2 - 2x - 2 = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

y = cosx + C

Differentiating both sides of this equation with respect to x, we get:

\begin{align}y^{'}=\frac{d}{dx}(cosx + C)\end{align}

=> y^' = - sinx

Substituting the value of y^'  in the given differential equation, we get:

L.H.S. = y^' + sinx = - sinx + sinx = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

\begin{align} y= \sqrt{1+x^2}\end{align}

Differentiating both sides of the equation with respect to x, we get:

\begin{align}  y^{'}=\frac{d}{dx}\left(\sqrt{1+x^2} \right)\end{align}

\begin{align}  y^{'}=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}\left(1+x^2\right)\end{align}

\begin{align}  y^{'}=\frac{2x}{2\sqrt{1+x^2}}\end{align}

\begin{align}  y^{'}=\frac{x}{\sqrt{1+x^2}}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{x}{\sqrt{1+x^2}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{\sqrt{1+x^2}}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{y}\end{align}

\begin{align}\Rightarrow y^{'}=\frac{xy}{1+x^2}\end{align}

∴ L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

y = Ax

Differentiating both sides with respect to x, we get: 

\begin{align}y^{'}=\frac{d}{dx}(Ax)\end{align}

⇒ y ^' = A

Substituting the value of y^' in the given differential equation, we get:

L.H.S. = xy^' = xA = Ax = y = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

y= x.sinx

Differentiating both sides of this equation with respect to x, we get:

\begin{align} y^{'} =\frac{d}{dx}\left(x.sinx\right)\end{align}

\begin{align}\Rightarrow y^{'} =sinx. \frac{d}{dx}\left(x\right)+ x. \frac{d}{dx}\left(sinx\right)\end{align}

\begin{align} \Rightarrow y^{'} =sinx + x.cosx\end{align}

Differentiating both sides of this equation with respect to x, we get:

L.H.S. =xy^' = x(sinx + xcosx)

\begin{align} =x.sinx + x^2.cosx\end{align}

\begin{align} =y + x^2.\sqrt{1-sin^2x}\end{align}

\begin{align} =y + x^2.\sqrt{1-\left(\frac{y}{x}\right)^2}\end{align}

\begin{align} =y + x^2.\sqrt{\frac{x^2-y^2}{x^2}}\end{align}

\begin{align} =y + x.\sqrt{x^2-y^2}\end{align}

R.H.S.