Ncert Solutions For Class 12 Maths Ex 9.2

Ncert Solutions For Class 12 Maths Chapter 9 Ex 9.2

Q.1: y = ex + 1; y” – y’ = 0

Solution:

y = ex + 1

Differentiate both the sides with respect to x, we get:

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

\(\Rightarrow\) y’ = ex . . . . . . . . . . . . . . (1)

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

\(\frac{d}{dx}(y^{‘})=\frac{d}{dx}(e^{x})\\\)

\(\\\boldsymbol{\rightarrow }\) 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.

 

 

Q.2: y = x2 + 2x + C; y’ – 2x – 2 = 0

Solution:

y = x2 + 2x + C

Differentiate both the sides with respect to x, we get:

\(y^{‘}=\frac{d}{dx}(x^{2}+2x+C)\\\)

\(\\\boldsymbol{\rightarrow }\) y’ = 2x + 2

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

y’ – 2x – 2 = 2x + 2 – 2x – 2 = 0 = R.H.S

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

Q.3: y = cos x + C; y’ + sin x = 0

Solution:

y = cos x + C

Differentiate both the sides with respect to x, we get:

y’ = \(\frac{d}{dx}\)(cos x + C )

\(\\\boldsymbol{\rightarrow }\) y’ = – sin x

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

y’ + sin x = -sin x + sin x = 0 = R.H.S.

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

 

 

Q.4: y = \(\sqrt{1+x^{2}}\); y’ = \(\frac{xy}{1+x^{2}}\)

Solution:

\(y=\sqrt{1+x^{2}}\)

Differentiate both the sides with respect to x, we get:

y’ = \(\frac{d}{dx}(\sqrt{1+x^{2}})\)

y’ = \(\frac{1}{2\sqrt{1+x^{2}}}.\frac{d}{dx}(1+x^{2})\)

y’ = \(\frac{2x}{2\sqrt{1+x^{2}}}\)

y’ = \(\frac{x}{\sqrt{1+x^{2}}}\)

y = \(\frac{x}{1+x^{2}}\times \sqrt{1+x^{2}}\)

y’ = \(\frac{x}{1+x^{2}}.y\)

y’ = \(\frac{xy}{1+x^{2}}\)

Therefore, L.H.S = R.H.S

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

 

 

Q.5: y = Ax; xy’ = y (x 0)

Solution:

Differentiate both the sides with respect to x, we get:

\(y^{‘}=\frac{d}{dx}(Ax)\\ \\ \Rightarrow y^{‘}=A\)

Substituting the values of \(y^{‘}\) in the given differential equation, we get the L.H.S. as:

\(xy^{‘}=x.A=Ax=y=R.H.S.\)

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

Q.6. \(y=x\sin x\;:\;xy^{‘}=y+x\sqrt{x^{2}-y^{2}}(x\neq 0\:and\:x>y\:or\:x<-y)\)

Solution:

y=x sin x

Differentiate both the sides with respect to x, we get:

\(y^{‘}=\frac{d}{dx}(x\sin x)\\ \\ \Rightarrow y^{‘}=\sin x.\frac{d}{dx}(x)+x.\frac{d}{dx}(\sin x)\\ \\ \Rightarrow y^{‘}=\sin x+x\cos x\)

Substitute the value of \(y^{‘}\) in the given differential equation, we get:

L.H.S = \(xy^{‘}=x(\sin x+x\cos x )\\ \\ =x\sin x+x^{2}\cos x \\ \\ =y+x^{2}.\sqrt{1-\sin ^{2}x}\\ \\ =y+x^{2}\sqrt{1-(\frac{y}{x})^{2}}\\ \\ =y+x\sqrt{y^{2}-x^{2}}\)=R.H.S.

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

Q.7. \(xy=log\, y+C\::\:y^{‘}=\frac{y^{2}}{1-xy}(xy\neq 1)\)

 

Solution:

\(xy=log \, y+C\)

Differentiate both the sides with respect to x, we get:

\(\frac{d}{dx}(xy)=\frac{d}{dx}(log \, y)\\ \\ \Rightarrow y.\frac{d}{dx}(x)+x.\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}\\ \\ \Rightarrow y+xy^{‘}=\frac{1}{y}y^{‘}\\ \\ \Rightarrow y^{2}+xy\,y^{‘}=y^{‘}\\ \\ \Rightarrow (xy-1)y^{‘}=-y^{2}\\ \\ \Rightarrow y^{‘}=\frac{y^{2}}{1-xy}\)

Therefore L.H.S = R.H.S

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

Q.8. \(y-\cos y=x\::\:(y\sin y+\cos y+x)y^{‘}=y\)

Solution:

\(y-\cos y=x\) ………..(1)

Differentiate both the sides with respect to x, we get:

\(\frac{dy}{dx}-\frac{d}{dx}(\cos y)=\frac{d}{dx}(x)\\ \\ \Rightarrow y^{‘}+\sin y.y^{‘}=1\\ \\ \Rightarrow y^{‘}(1+\sin y)=1\\ \\ \Rightarrow y^{‘}=\frac{1}{1+\sin y}\)

Substitute the value of \(y^{‘}\) in the given differential equation, we get:

L.H.S = \((y\sin y+\cos y+x)y^{‘}\\ \\ =(y\sin y+\cos y+y-\cos y)\times \frac{1}{1+\sin y}\\ \\ =y(1+\sin y).\frac{1}{1+\sin y}\\ \\ =y\\ \\ =R.H.S.\)

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

 

Q.9: \(x+y=\tan ^{-1}y\::\:y^{2}y^{‘}+y^{2}+1=0\)

 

Solution: \(x+y=\tan ^{-1}y\)

Differentiate both the sides with respect to x, we get:

\(\frac{d}{dx}(x+y)=\frac{d}{dx}(\tan ^{-1}y)\\ \\ \Rightarrow 1+y^{‘}=[\frac{1}{1+y^{2}}]y^{‘}\\ \\ \Rightarrow y^{‘}[\frac{1}{1+y^{2}}-1]=1\\ \\ \Rightarrow y^{‘}[\frac{1-(1+y^{2})}{1+y^{2}}]=1\\ \\ \Rightarrow y^{‘}[\frac{-y^{2}}{1+y^{2}}]=1\\ \\ \Rightarrow y^{‘}=\frac{-(1+y^{2})}{y^{2}}\)

Substitute the value of \(y^{‘}\) in the given differential equation, we get:

L.H.S = \(y^{2}y^{‘}+y^{2}+1=y^{2}[\frac{-(1+y^{2})}{y^{2}}]+y^{2}+1\\ \\ =-1-y^{2}+y^{2}+1\\ \\ =0\\ \\ =R.H.S.\)

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

Q.10: \(y=\sqrt{a^{2}-x^{2}}x\in (-a,a)\: :\: x+y\frac{dy}{dx}=0(y\neq 0)\)

Solution:

\(y=\sqrt{a^{2}-x^{2}}\)

Differentiate both the sides with respect to x, we get:

\(\frac{dy}{dx}=\frac{d}{dx}(\sqrt{a^{2}-x^{2}})\\ \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{a^{2}-x^{2}}}.\frac{d}{dx}(a^{2}-x^{2})\\ \\ =\frac{1}{2\sqrt{a^{2}-x^{2}}}(-2x)\\ \\ =\frac{-x}{\sqrt{a^{2}-x^{2}}}\)

Substitute the value of \(\frac{dy}{dx}\)in the given differential equation, we get:

L.H.S = \(x+y\frac{dy}{dx}=x+\sqrt{a^{2}-x^{2}}\times \frac{-x}{\sqrt{a^{2}-x^{2}}}\\ \\ =x-x\\ \\ =0\\ \\ R.H.S.\)

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

 

 

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

(i) 0

(ii) 2

(iii) 3

(iv) 4

Solution:

We know that, number of constants in the general solution of a differential equation of order n is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Therefore, (iv) is the correct answer.

 

 

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

(i) 3

(ii) 2

(iii) 1

(iv) 0

Solution:

In a particular solution of a differential equation, there are no arbitrary constants. Hence, the correct answer is (iv).

 

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