NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2

Q.1

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ {\text{y = e}}^{\text{x}}\text{+ 1}\text{:}\text{y” – y’ = 0}\end{array}

Ans

\begin{array}{l}{\text{y = e}}^{\text{x}}\text{+ 1}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{y’ = e}}^{\text{x}}\mathrm{\dots }\left(\text{i}\right)\\ \text{Differentiating equation}\left(\text{i}\right)\text{w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{y” = e}}^{\text{x}}\\ \text{Substituting values of y’ and y” in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y” – y’ = 0, we get}\\ {\text{y” – y’ =e}}^{\text{x}}{\text{– e}}^{\text{x}}\text{= 0 = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.2

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ {\text{y = x}}^{\text{2}}\text{+2x + C}\text{:}\text{y’ –2x –2 = 0}\end{array}

Ans

\begin{array}{l}{\text{y = x}}^{\text{2}}\text{+2x + C}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ = 2x + 2}\\ \text{Substituting values of y’ in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y’ –2x –2 = 0, we get}\\ \text{y’ – 2x –2 = 2x + 2 –2x –2}\\ \text{= 0 = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.3

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:} \; \\ \text{y = cos x + C}\text{:}\text{y’ + sinx = 0}\end{array}

Ans

\begin{array}{l}\text{y = cos x + C}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ = –sinx}\\ \text{Substituting values of y’ in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y’ + sinx = 0, we get}\\ \text{y’ + sinx = –sinx + sinx = 0 = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.4

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ \text{y =}\sqrt{{\text{1 + x}}^{\text{2}}}\text{:}\text{y’ =}\frac{\text{xy}}{{\text{1 + x}}^{\text{2}}}\end{array}

Ans

\begin{array}{l}\text{y =}\sqrt{{\text{1 + x}}^{\text{2}}}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ =}\frac{\text{x}}{\sqrt{{\text{1 + x}}^{\text{2}}}}\\ \text{Substituting values of y in R}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y’ =}\frac{\text{xy}}{{\text{1 + x}}^{\text{2}}}\text{, we get}\\ \text{R}\text{.H}\text{.S}\text{. =}\frac{\text{xy}}{{\text{1 + x}}^{\text{2}}}\\ \text{=}\frac{\text{x}\left(\sqrt{{\text{1 + x}}^{\text{2}}}\right)}{{\text{1 + x}}^{\text{2}}}\\ \text{=}\frac{\text{x}}{\sqrt{{\text{1 + x}}^{\text{2}}}}\\ \text{= y’ = L}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.5

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ \text{y = Ax}\text{:}\text{xy’ = y}\left(\text{x}\ne \text{0}\right)\end{array}

Ans

\begin{array}{l}\text{y = Ax}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ = A}\\ \text{Substituting values of y’ in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation xy’ = y, we get}\\ \text{L}\text{.H}\text{.S}\text{. = xy’}\\ \text{= x}\left(\text{A}\right)\\ \text{= Ax = y = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.6

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ \text{y = x sinx}\text{: xy’ = y + x}\sqrt{{\text{x}}^{\text{2}}{\text{– y}}^{\text{2}}}\left(\text{x}\ne \text{0}\text{and x > y or x < –y}\right)\end{array}

Ans

\begin{array}{l}\text{y = xsin x}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ = xcos x + sinx}\\ \text{Substituting values of y’ in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation xy’ = y + x}\sqrt{{\text{x}}^{\text{2}}{\text{– y}}^{\text{2}}}\text{, we get}\\ \text{L}\text{.H}\text{.S}\text{. = xy’ = x}\left(\text{xcosx + sinx}\right)\\ {\text{= x}}^{\text{2}}\text{cosx + xsinx}\\ \text{R}\text{.H}\text{.S}\text{. = xsinx + x}\sqrt{{\text{x}}^{\text{2}}\text{–}{\left(\text{xsinx}\right)}^{\text{2}}}\\ {\text{= xsinx + x}}^{\text{2}}\sqrt{{\text{1 – sin}}^{\text{2}}\text{x}}\\ {\text{= xsinx + x}}^{\text{2}}\text{cosx}\\ \text{So, L}\text{.H}\text{.S}\text{. = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.7

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ \text{xy = logy + C : y’ =}\frac{{\text{y}}^{\text{2}}}{\text{1 – xy}}\left(\text{xy}\ne \text{0}\right)\end{array}

Ans

\begin{array}{l}\text{xy = logy + C}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{xy’ + y =}\frac{\text{1}}{\text{y}}\text{y’ + 0}\\ {\text{xyy’ + y}}^{\text{2}}\text{= y’}\\ {\text{y}}^{\text{2}}\text{= y’ – xyy’}\\ \text{= y’}\left(\text{1 – xy}\right)\\ \text{y’ =}\frac{{\text{y}}^{\text{2}}}{\left(\text{1 – xy}\right)}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.8

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ \text{y – cosy = x :}\left(\text{ysiny + cosy + x}\right)\text{y’ = y}\end{array}

Ans

\begin{array}{l}y-\cos y=x\mathrm{\dots }\left(i\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’}+\sin y.y‘=1\\ y‘\left(1+\sin y\right)=1\\ y‘=\frac{1}{1+\sin y}\\ \text{Substituting values of y’ in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation}\left(\text{ysiny + cosy + x}\right)\text{y’ = y, we get}\\ \left(y\sin y+\cos y+x\right)y‘=\left(y\sin y+\cos y+x\right)\frac{1}{\left(1+\sin y\right)}\\ =\left(y\sin y+\cos y+y-\cos y\right)\frac{1}{\left(1+\sin y\right)}\\ \left[\text{From equation}\left(\text{i}\right)\right]\\ =\left(y\sin y+y\right)\frac{1}{\left(1+\sin y\right)}\\ =y\left(\sin y+1\right)\frac{1}{\left(1+\sin y\right)}\\ =y=R.H.S.\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.9

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ x+y=ta{n}^{–1}y:y^{2}y‘+y^2+1=0\end{array}

Ans

\begin{array}{l}x+y={\tan }^{–1}y\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{1+y’}=\left(\frac{1}{1+y^2}\right)y‘\\ \left(\text{1}+\text{y’}\right)\left(1+y^2\right)=y‘\\ 1+y^2+y‘+y‘y^2=y‘\\ 1+y^2+y‘y^2=0\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.10

\begin{array}{l} \text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ y=\sqrt{a^2-x^2}x\in \left(-a,a\right):x+y\frac{dy}{dx}=0\left(y\ne 0\right)\end{array}

Ans

\begin{array}{l}y=\sqrt{a^2-x^2}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}\\ =\frac{-x}{y}\\ \text{Substituting the value of}\frac{dy}{dx}\text{in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation}x+y\frac{dy}{dx}=0,\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x+y\frac{dy}{dx}\\ =x+y\left(\frac{-x}{y}\right)\\ =x-x\\ =0=R.H.S.\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.11  The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Ans

Since, number of constants in a differential equation of order n is equal to its order i.e., n.
Thus, number of arbitrary constants in a differential equation of fourth order are 4.
Thus, option D is correct.

Q.12  The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Ans

In a particular solution, there are no arbitrary constants.

∴ Number of arbitrary constants = 0
Thus, option D is correct.