87301
The solution of \(\tan \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})\) is
1 \(\sec y=2 \cos x+c\)
2 \(\sec y=-2 \cos x+c\)
3 \(\tan y=-2 \cos x+c\)
4 \(\sec ^{2} y=-2 \cos x+c\)
Explanation:
(B) : \(\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})}{\tan \mathrm{y}}\) \(\left(\frac{d y}{d x}\right)=\frac{2 \sin x \cos y}{\tan y}\) \(\frac{\tan y}{\cos y} d y=2 \sin d x\) \(\frac{\sin y}{\cos ^{2} y} d y=2 \sin d x\) \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin d x\) \(\frac{1}{\cos y}=c-2 \cos x\) \(\sec y=-2 \cos x+c\)
AP EAMCET-2010
Differential Equation
87302
The solution of the differential equation \(\frac{d y}{d x}\) \(\boldsymbol{\operatorname { s i n }}(\mathrm{x}+\mathrm{y}) \boldsymbol{\operatorname { t a n }}(\mathrm{x}+\mathrm{y})-\mathbf{1}\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+c\)
2 \(x+\operatorname{cosec}(x+y)=c\)
3 \(x+\tan (x+y)=c\)
4 \(x+\sec (x+y)=c\)
Explanation:
(B) : We have given, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1 \tag{i}\) Put, \(x+y=v \Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) From equation (i) \(\frac{d v}{d x}-1=\sin v \tan v-1 \Rightarrow \frac{d v}{d x}=\sin v \tan v\) \(\frac{1}{\sin v \tan v} d v=d x\) \(\operatorname{cosec} v \cdot \cot v d v=d x\) On integrating both side we get- \(\int \operatorname{cosec} v \cdot \cot v d v=\int d x\) \(-\operatorname{cosec} v=x+c\) \(x+\operatorname{cosec} v=c\) Putting, \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get- \(x+\operatorname{cosec}(x+y)=c\)
AP EAMCET-2009
Differential Equation
87303
The solution of \(\frac{d x}{d y}+\frac{x}{y}=x^{2}\) is:
(A) : Given, \(\left(\frac{d y}{d x}\right)+1=e^{x+y}\) Let, \(x+y=z\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}+1=\frac{\mathrm{dz}}{\mathrm{dx}}\) Now, comparing (i) and (ii) we get, \(\frac{\mathrm{dz}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{z}} \Rightarrow \mathrm{dx}=\mathrm{e}^{-\mathrm{z}} \mathrm{dz}\) On integrating both sides, we get - \(\int \mathrm{dx}=\int \mathrm{e}^{-\mathrm{z}} \mathrm{dz} \Rightarrow \mathrm{x}+\mathrm{c}=-\mathrm{e}^{-\mathrm{z}}\) \(\mathrm{x}+\mathrm{e}^{-(\mathrm{x}+\mathrm{y})}+\mathrm{c}=0\)
AP EAMCET-2007
Differential Equation
87306
An integrating facto of the equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0 \text { is }\)
1 \(e^{x}\)
2 \(x^{2}\)
3 \(\frac{1}{\mathrm{x}}\)
4 \(x\)
Explanation:
(D) : Dividing the entire equation by 'dx' gives as \(\left(1+y+x^{2} y\right)+\frac{d y}{d x}\left(x+x^{3}\right)=0\) \(\frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x+x^{3}}=\frac{-1}{x+x^{3}} \Rightarrow \frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=\frac{-1}{x+x^{3}}\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{\left.x+x^{3} x^{3}\right) d x} \frac{d y}{d x}+y P(x)=Q(x)\) Hence, I.F. \(=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}\) \(\mathrm{e}^{\int \frac{1}{x} d x}\)
87301
The solution of \(\tan \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})\) is
1 \(\sec y=2 \cos x+c\)
2 \(\sec y=-2 \cos x+c\)
3 \(\tan y=-2 \cos x+c\)
4 \(\sec ^{2} y=-2 \cos x+c\)
Explanation:
(B) : \(\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})}{\tan \mathrm{y}}\) \(\left(\frac{d y}{d x}\right)=\frac{2 \sin x \cos y}{\tan y}\) \(\frac{\tan y}{\cos y} d y=2 \sin d x\) \(\frac{\sin y}{\cos ^{2} y} d y=2 \sin d x\) \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin d x\) \(\frac{1}{\cos y}=c-2 \cos x\) \(\sec y=-2 \cos x+c\)
AP EAMCET-2010
Differential Equation
87302
The solution of the differential equation \(\frac{d y}{d x}\) \(\boldsymbol{\operatorname { s i n }}(\mathrm{x}+\mathrm{y}) \boldsymbol{\operatorname { t a n }}(\mathrm{x}+\mathrm{y})-\mathbf{1}\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+c\)
2 \(x+\operatorname{cosec}(x+y)=c\)
3 \(x+\tan (x+y)=c\)
4 \(x+\sec (x+y)=c\)
Explanation:
(B) : We have given, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1 \tag{i}\) Put, \(x+y=v \Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) From equation (i) \(\frac{d v}{d x}-1=\sin v \tan v-1 \Rightarrow \frac{d v}{d x}=\sin v \tan v\) \(\frac{1}{\sin v \tan v} d v=d x\) \(\operatorname{cosec} v \cdot \cot v d v=d x\) On integrating both side we get- \(\int \operatorname{cosec} v \cdot \cot v d v=\int d x\) \(-\operatorname{cosec} v=x+c\) \(x+\operatorname{cosec} v=c\) Putting, \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get- \(x+\operatorname{cosec}(x+y)=c\)
AP EAMCET-2009
Differential Equation
87303
The solution of \(\frac{d x}{d y}+\frac{x}{y}=x^{2}\) is:
(A) : Given, \(\left(\frac{d y}{d x}\right)+1=e^{x+y}\) Let, \(x+y=z\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}+1=\frac{\mathrm{dz}}{\mathrm{dx}}\) Now, comparing (i) and (ii) we get, \(\frac{\mathrm{dz}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{z}} \Rightarrow \mathrm{dx}=\mathrm{e}^{-\mathrm{z}} \mathrm{dz}\) On integrating both sides, we get - \(\int \mathrm{dx}=\int \mathrm{e}^{-\mathrm{z}} \mathrm{dz} \Rightarrow \mathrm{x}+\mathrm{c}=-\mathrm{e}^{-\mathrm{z}}\) \(\mathrm{x}+\mathrm{e}^{-(\mathrm{x}+\mathrm{y})}+\mathrm{c}=0\)
AP EAMCET-2007
Differential Equation
87306
An integrating facto of the equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0 \text { is }\)
1 \(e^{x}\)
2 \(x^{2}\)
3 \(\frac{1}{\mathrm{x}}\)
4 \(x\)
Explanation:
(D) : Dividing the entire equation by 'dx' gives as \(\left(1+y+x^{2} y\right)+\frac{d y}{d x}\left(x+x^{3}\right)=0\) \(\frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x+x^{3}}=\frac{-1}{x+x^{3}} \Rightarrow \frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=\frac{-1}{x+x^{3}}\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{\left.x+x^{3} x^{3}\right) d x} \frac{d y}{d x}+y P(x)=Q(x)\) Hence, I.F. \(=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}\) \(\mathrm{e}^{\int \frac{1}{x} d x}\)
87301
The solution of \(\tan \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})\) is
1 \(\sec y=2 \cos x+c\)
2 \(\sec y=-2 \cos x+c\)
3 \(\tan y=-2 \cos x+c\)
4 \(\sec ^{2} y=-2 \cos x+c\)
Explanation:
(B) : \(\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})}{\tan \mathrm{y}}\) \(\left(\frac{d y}{d x}\right)=\frac{2 \sin x \cos y}{\tan y}\) \(\frac{\tan y}{\cos y} d y=2 \sin d x\) \(\frac{\sin y}{\cos ^{2} y} d y=2 \sin d x\) \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin d x\) \(\frac{1}{\cos y}=c-2 \cos x\) \(\sec y=-2 \cos x+c\)
AP EAMCET-2010
Differential Equation
87302
The solution of the differential equation \(\frac{d y}{d x}\) \(\boldsymbol{\operatorname { s i n }}(\mathrm{x}+\mathrm{y}) \boldsymbol{\operatorname { t a n }}(\mathrm{x}+\mathrm{y})-\mathbf{1}\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+c\)
2 \(x+\operatorname{cosec}(x+y)=c\)
3 \(x+\tan (x+y)=c\)
4 \(x+\sec (x+y)=c\)
Explanation:
(B) : We have given, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1 \tag{i}\) Put, \(x+y=v \Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) From equation (i) \(\frac{d v}{d x}-1=\sin v \tan v-1 \Rightarrow \frac{d v}{d x}=\sin v \tan v\) \(\frac{1}{\sin v \tan v} d v=d x\) \(\operatorname{cosec} v \cdot \cot v d v=d x\) On integrating both side we get- \(\int \operatorname{cosec} v \cdot \cot v d v=\int d x\) \(-\operatorname{cosec} v=x+c\) \(x+\operatorname{cosec} v=c\) Putting, \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get- \(x+\operatorname{cosec}(x+y)=c\)
AP EAMCET-2009
Differential Equation
87303
The solution of \(\frac{d x}{d y}+\frac{x}{y}=x^{2}\) is:
(A) : Given, \(\left(\frac{d y}{d x}\right)+1=e^{x+y}\) Let, \(x+y=z\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}+1=\frac{\mathrm{dz}}{\mathrm{dx}}\) Now, comparing (i) and (ii) we get, \(\frac{\mathrm{dz}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{z}} \Rightarrow \mathrm{dx}=\mathrm{e}^{-\mathrm{z}} \mathrm{dz}\) On integrating both sides, we get - \(\int \mathrm{dx}=\int \mathrm{e}^{-\mathrm{z}} \mathrm{dz} \Rightarrow \mathrm{x}+\mathrm{c}=-\mathrm{e}^{-\mathrm{z}}\) \(\mathrm{x}+\mathrm{e}^{-(\mathrm{x}+\mathrm{y})}+\mathrm{c}=0\)
AP EAMCET-2007
Differential Equation
87306
An integrating facto of the equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0 \text { is }\)
1 \(e^{x}\)
2 \(x^{2}\)
3 \(\frac{1}{\mathrm{x}}\)
4 \(x\)
Explanation:
(D) : Dividing the entire equation by 'dx' gives as \(\left(1+y+x^{2} y\right)+\frac{d y}{d x}\left(x+x^{3}\right)=0\) \(\frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x+x^{3}}=\frac{-1}{x+x^{3}} \Rightarrow \frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=\frac{-1}{x+x^{3}}\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{\left.x+x^{3} x^{3}\right) d x} \frac{d y}{d x}+y P(x)=Q(x)\) Hence, I.F. \(=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}\) \(\mathrm{e}^{\int \frac{1}{x} d x}\)
87301
The solution of \(\tan \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})\) is
1 \(\sec y=2 \cos x+c\)
2 \(\sec y=-2 \cos x+c\)
3 \(\tan y=-2 \cos x+c\)
4 \(\sec ^{2} y=-2 \cos x+c\)
Explanation:
(B) : \(\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})}{\tan \mathrm{y}}\) \(\left(\frac{d y}{d x}\right)=\frac{2 \sin x \cos y}{\tan y}\) \(\frac{\tan y}{\cos y} d y=2 \sin d x\) \(\frac{\sin y}{\cos ^{2} y} d y=2 \sin d x\) \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin d x\) \(\frac{1}{\cos y}=c-2 \cos x\) \(\sec y=-2 \cos x+c\)
AP EAMCET-2010
Differential Equation
87302
The solution of the differential equation \(\frac{d y}{d x}\) \(\boldsymbol{\operatorname { s i n }}(\mathrm{x}+\mathrm{y}) \boldsymbol{\operatorname { t a n }}(\mathrm{x}+\mathrm{y})-\mathbf{1}\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+c\)
2 \(x+\operatorname{cosec}(x+y)=c\)
3 \(x+\tan (x+y)=c\)
4 \(x+\sec (x+y)=c\)
Explanation:
(B) : We have given, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1 \tag{i}\) Put, \(x+y=v \Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) From equation (i) \(\frac{d v}{d x}-1=\sin v \tan v-1 \Rightarrow \frac{d v}{d x}=\sin v \tan v\) \(\frac{1}{\sin v \tan v} d v=d x\) \(\operatorname{cosec} v \cdot \cot v d v=d x\) On integrating both side we get- \(\int \operatorname{cosec} v \cdot \cot v d v=\int d x\) \(-\operatorname{cosec} v=x+c\) \(x+\operatorname{cosec} v=c\) Putting, \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get- \(x+\operatorname{cosec}(x+y)=c\)
AP EAMCET-2009
Differential Equation
87303
The solution of \(\frac{d x}{d y}+\frac{x}{y}=x^{2}\) is:
(A) : Given, \(\left(\frac{d y}{d x}\right)+1=e^{x+y}\) Let, \(x+y=z\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}+1=\frac{\mathrm{dz}}{\mathrm{dx}}\) Now, comparing (i) and (ii) we get, \(\frac{\mathrm{dz}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{z}} \Rightarrow \mathrm{dx}=\mathrm{e}^{-\mathrm{z}} \mathrm{dz}\) On integrating both sides, we get - \(\int \mathrm{dx}=\int \mathrm{e}^{-\mathrm{z}} \mathrm{dz} \Rightarrow \mathrm{x}+\mathrm{c}=-\mathrm{e}^{-\mathrm{z}}\) \(\mathrm{x}+\mathrm{e}^{-(\mathrm{x}+\mathrm{y})}+\mathrm{c}=0\)
AP EAMCET-2007
Differential Equation
87306
An integrating facto of the equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0 \text { is }\)
1 \(e^{x}\)
2 \(x^{2}\)
3 \(\frac{1}{\mathrm{x}}\)
4 \(x\)
Explanation:
(D) : Dividing the entire equation by 'dx' gives as \(\left(1+y+x^{2} y\right)+\frac{d y}{d x}\left(x+x^{3}\right)=0\) \(\frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x+x^{3}}=\frac{-1}{x+x^{3}} \Rightarrow \frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=\frac{-1}{x+x^{3}}\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{\left.x+x^{3} x^{3}\right) d x} \frac{d y}{d x}+y P(x)=Q(x)\) Hence, I.F. \(=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}\) \(\mathrm{e}^{\int \frac{1}{x} d x}\)
87301
The solution of \(\tan \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})\) is
1 \(\sec y=2 \cos x+c\)
2 \(\sec y=-2 \cos x+c\)
3 \(\tan y=-2 \cos x+c\)
4 \(\sec ^{2} y=-2 \cos x+c\)
Explanation:
(B) : \(\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})}{\tan \mathrm{y}}\) \(\left(\frac{d y}{d x}\right)=\frac{2 \sin x \cos y}{\tan y}\) \(\frac{\tan y}{\cos y} d y=2 \sin d x\) \(\frac{\sin y}{\cos ^{2} y} d y=2 \sin d x\) \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin d x\) \(\frac{1}{\cos y}=c-2 \cos x\) \(\sec y=-2 \cos x+c\)
AP EAMCET-2010
Differential Equation
87302
The solution of the differential equation \(\frac{d y}{d x}\) \(\boldsymbol{\operatorname { s i n }}(\mathrm{x}+\mathrm{y}) \boldsymbol{\operatorname { t a n }}(\mathrm{x}+\mathrm{y})-\mathbf{1}\) is
1 \(\operatorname{cosec}(x+y)+\tan (x+y)=x+c\)
2 \(x+\operatorname{cosec}(x+y)=c\)
3 \(x+\tan (x+y)=c\)
4 \(x+\sec (x+y)=c\)
Explanation:
(B) : We have given, \(\frac{d y}{d x}=\sin (x+y) \tan (x+y)-1 \tag{i}\) Put, \(x+y=v \Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x}\) \(\frac{d y}{d x}=\frac{d v}{d x}-1\) From equation (i) \(\frac{d v}{d x}-1=\sin v \tan v-1 \Rightarrow \frac{d v}{d x}=\sin v \tan v\) \(\frac{1}{\sin v \tan v} d v=d x\) \(\operatorname{cosec} v \cdot \cot v d v=d x\) On integrating both side we get- \(\int \operatorname{cosec} v \cdot \cot v d v=\int d x\) \(-\operatorname{cosec} v=x+c\) \(x+\operatorname{cosec} v=c\) Putting, \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get- \(x+\operatorname{cosec}(x+y)=c\)
AP EAMCET-2009
Differential Equation
87303
The solution of \(\frac{d x}{d y}+\frac{x}{y}=x^{2}\) is:
(A) : Given, \(\left(\frac{d y}{d x}\right)+1=e^{x+y}\) Let, \(x+y=z\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}+1=\frac{\mathrm{dz}}{\mathrm{dx}}\) Now, comparing (i) and (ii) we get, \(\frac{\mathrm{dz}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{z}} \Rightarrow \mathrm{dx}=\mathrm{e}^{-\mathrm{z}} \mathrm{dz}\) On integrating both sides, we get - \(\int \mathrm{dx}=\int \mathrm{e}^{-\mathrm{z}} \mathrm{dz} \Rightarrow \mathrm{x}+\mathrm{c}=-\mathrm{e}^{-\mathrm{z}}\) \(\mathrm{x}+\mathrm{e}^{-(\mathrm{x}+\mathrm{y})}+\mathrm{c}=0\)
AP EAMCET-2007
Differential Equation
87306
An integrating facto of the equation \(\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0 \text { is }\)
1 \(e^{x}\)
2 \(x^{2}\)
3 \(\frac{1}{\mathrm{x}}\)
4 \(x\)
Explanation:
(D) : Dividing the entire equation by 'dx' gives as \(\left(1+y+x^{2} y\right)+\frac{d y}{d x}\left(x+x^{3}\right)=0\) \(\frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x+x^{3}}=\frac{-1}{x+x^{3}} \Rightarrow \frac{d y}{d x}+\frac{y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)}=\frac{-1}{x+x^{3}}\) \(\frac{d y}{d x}+\frac{y}{x}=-\frac{1}{\left.x+x^{3} x^{3}\right) d x} \frac{d y}{d x}+y P(x)=Q(x)\) Hence, I.F. \(=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}\) \(\mathrm{e}^{\int \frac{1}{x} d x}\)
