# Chapter 9 Differential Equations

In this chapter, we will shidy about differential equations and their solutions. Concepts of differential equations will help us in this class and also in higher studies. It has many applications in other branches like science, algebra, etc. Key topics of this chapter are - definition, order and degree, general and particular solutions, formation of differential equaňon whose general solution is given, method of separation of variables, homogeneous differential equations of first order and degree.

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### Exercise 1

•  Q1 Determine order and degree(if defined) of differential equation \begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0\end{align} Ans: \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 ym '. Therefore, its order is four. The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined. Q2 Determine order and degree(if defined) of differential equation y' + 5y = 0 Ans: 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. Q3 Determine order and degree(if defined) of differential equation \begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align} Ans: \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} Q4 Determine order and degree(if defined) of differential equation \begin{align}\left(\frac{d^2y}{dx^2}\right)^2\;+\;cos\left(\frac{dy}{dx}\right)\;=\;0\end{align} Ans: \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. Q5 Determine order and degree(if defined) of differential equation \begin{align}\frac{d^2y}{dx^2}=\cos3x  + sin3x\end{align} Ans: \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. Q6 Determine order and degree(if defined) of differential equation (ym)2 + (yn)3 + (y')4 + y5 =0 Ans: (ym)2 + (yn)3 + (y')4 + y5 =0 The highest order derivative present in the differential equation isym. Therefore, its order is three. The given differential equation is a polynomial equation in ym ,  yn , y'. The highest power raised to ym is 2. Hence, its degree is 2. Q7 Determine order and degree(if defined) of differential equation ym + 2yn + y' =0 Ans: The highest order derivative present in the differential equation is ym. Therefore, its order is three. It is a polynomial equation in ym , yn  and y' . The highest power raised to ym is 1. Hence, its degree is 1. Q8 Determine order and degree(if defined) of differential y' + y =ex Ans: y' + y =ex y' + y  - ex =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. Q9 Determine order and degree(if defined) of differential equation yn + (y')2 + 2y =0 Ans: 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. Q10 Determine order and degree(if defined) of differential equation yn + 2y' + siny = 0 Ans: 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. Q11 The degree of the differential equation \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} is (A) 3 (B) 2 (C) 1 (D) not defined Ans: \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. Q12 The order of the differential equation \begin{align}2x^2\frac{d^2y}{dx^2}\;- \;3\frac{dy}{dx}\;+ y=\;0\end{align} is (A) 2 (B) 1 (C) 0 (D) not defined Ans: \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.

### Exercise 2

•  Q1 y = ex +1 : yn -y' = 0 Ans: y = ex +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' = ex                          ...(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'' = ex Substituting the values of y' and y'' in the given differential equation, we get the L.H.S. as: y'' - y' = ex - ex = 0 = R.H.S. Thus, the given function is the solution of the corresponding differential equation. Q2 y = x2 + 2x + C : y' - 2x - 2 = 0 Ans: 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. Q3 y = cosx + C : y' + sinx = 0 Ans: 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. Q4 \begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align} Ans: \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. Q5 y = Ax : xy' = y (x ≠ 0) Ans: 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. Q6 \begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align} Ans: 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.