87571
The solution of the differential equation \(\frac{d y}{d x}=\frac{1}{x+y^{2}}\) is
1 \(y=-x^{2}-2 x-2+c^{x}\)
2 \(y=x^{2}+2 x+2-c e^{x}\)
3 \(x=-y^{2}-2 y+2-c e^{y}\)
4 \(x=-y^{2}-2 y-2+c e^{y}\)
5 \(x=y^{2}+2 y+2-c e^{y}\)
Explanation:
(D) : Given differential equation, \(\frac{d y}{d x}=\frac{1}{x+y^{2}} \Rightarrow \frac{d x}{d y}=\frac{x+y^{2}}{1} \Rightarrow \frac{d x}{d y}-x=y^{2}\) Assume, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{y}^{2}\) I.F, \(=e \int-1 d y=e^{-y}\) \(x e^{-y}=\int e^{-y} y^{2} d y=-e^{-y} y^{2}+\int 2 e^{-y} y d y\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y+\int e^{-y} d y\right]+c\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y-e^{-y}\right]+c\) \(\Rightarrow x e^{-y}=e^{-y}\left(-y^{2}-2 y-2\right)+c\) \(x=-y^{2}-2 y-2+c e^{y}\)
Kerala CEE-2009
Differential Equation
87572
The solution of \(\frac{d y}{d x}+y \tan x=\sec x\) is :
1 \(y \sec x=\tan x+c\)
2 \(y \tan x=\sec x+c\)
3 \(\tan x=y \tan x+c\)
4 \(x \sec x=\tan y+c\)
5 \(x \tan x=y \tan x+c\)
Explanation:
(A) : Given, \(\frac{d y}{d x}+y \tan x=\sec x\) It is a first-degree linear D.E. of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\theta\) Here, \(\mathrm{P}=\tan \mathrm{x}, \theta=\sec \mathrm{x}\) I.F. \(=\mathrm{e}^{\int \operatorname{pdx}}=\mathrm{e}^{\int \tan x \mathrm{~d} x}=\mathrm{e}^{\log \sec \mathrm{x}}=\sec \mathrm{x}\) The solution is given by- \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F} \cdot=\int \mathrm{Q} \cdot \mathrm{I} \cdot \mathrm{F} \cdot \mathrm{dx}+\mathrm{c}\) \(y \cdot \sec x=\int \sec ^{2} x d x+c\) \(\mathrm{y} \cdot \sec \mathrm{x}=\tan \mathrm{x}+\mathrm{c}\)
Kerala CEE-2004
Differential Equation
87573
Solution of the differential equation \(x=1+x y \frac{d y}{d x}+\frac{(x y)^{2}}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{(x y)^{3}}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is
87571
The solution of the differential equation \(\frac{d y}{d x}=\frac{1}{x+y^{2}}\) is
1 \(y=-x^{2}-2 x-2+c^{x}\)
2 \(y=x^{2}+2 x+2-c e^{x}\)
3 \(x=-y^{2}-2 y+2-c e^{y}\)
4 \(x=-y^{2}-2 y-2+c e^{y}\)
5 \(x=y^{2}+2 y+2-c e^{y}\)
Explanation:
(D) : Given differential equation, \(\frac{d y}{d x}=\frac{1}{x+y^{2}} \Rightarrow \frac{d x}{d y}=\frac{x+y^{2}}{1} \Rightarrow \frac{d x}{d y}-x=y^{2}\) Assume, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{y}^{2}\) I.F, \(=e \int-1 d y=e^{-y}\) \(x e^{-y}=\int e^{-y} y^{2} d y=-e^{-y} y^{2}+\int 2 e^{-y} y d y\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y+\int e^{-y} d y\right]+c\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y-e^{-y}\right]+c\) \(\Rightarrow x e^{-y}=e^{-y}\left(-y^{2}-2 y-2\right)+c\) \(x=-y^{2}-2 y-2+c e^{y}\)
Kerala CEE-2009
Differential Equation
87572
The solution of \(\frac{d y}{d x}+y \tan x=\sec x\) is :
1 \(y \sec x=\tan x+c\)
2 \(y \tan x=\sec x+c\)
3 \(\tan x=y \tan x+c\)
4 \(x \sec x=\tan y+c\)
5 \(x \tan x=y \tan x+c\)
Explanation:
(A) : Given, \(\frac{d y}{d x}+y \tan x=\sec x\) It is a first-degree linear D.E. of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\theta\) Here, \(\mathrm{P}=\tan \mathrm{x}, \theta=\sec \mathrm{x}\) I.F. \(=\mathrm{e}^{\int \operatorname{pdx}}=\mathrm{e}^{\int \tan x \mathrm{~d} x}=\mathrm{e}^{\log \sec \mathrm{x}}=\sec \mathrm{x}\) The solution is given by- \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F} \cdot=\int \mathrm{Q} \cdot \mathrm{I} \cdot \mathrm{F} \cdot \mathrm{dx}+\mathrm{c}\) \(y \cdot \sec x=\int \sec ^{2} x d x+c\) \(\mathrm{y} \cdot \sec \mathrm{x}=\tan \mathrm{x}+\mathrm{c}\)
Kerala CEE-2004
Differential Equation
87573
Solution of the differential equation \(x=1+x y \frac{d y}{d x}+\frac{(x y)^{2}}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{(x y)^{3}}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is
87571
The solution of the differential equation \(\frac{d y}{d x}=\frac{1}{x+y^{2}}\) is
1 \(y=-x^{2}-2 x-2+c^{x}\)
2 \(y=x^{2}+2 x+2-c e^{x}\)
3 \(x=-y^{2}-2 y+2-c e^{y}\)
4 \(x=-y^{2}-2 y-2+c e^{y}\)
5 \(x=y^{2}+2 y+2-c e^{y}\)
Explanation:
(D) : Given differential equation, \(\frac{d y}{d x}=\frac{1}{x+y^{2}} \Rightarrow \frac{d x}{d y}=\frac{x+y^{2}}{1} \Rightarrow \frac{d x}{d y}-x=y^{2}\) Assume, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{y}^{2}\) I.F, \(=e \int-1 d y=e^{-y}\) \(x e^{-y}=\int e^{-y} y^{2} d y=-e^{-y} y^{2}+\int 2 e^{-y} y d y\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y+\int e^{-y} d y\right]+c\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y-e^{-y}\right]+c\) \(\Rightarrow x e^{-y}=e^{-y}\left(-y^{2}-2 y-2\right)+c\) \(x=-y^{2}-2 y-2+c e^{y}\)
Kerala CEE-2009
Differential Equation
87572
The solution of \(\frac{d y}{d x}+y \tan x=\sec x\) is :
1 \(y \sec x=\tan x+c\)
2 \(y \tan x=\sec x+c\)
3 \(\tan x=y \tan x+c\)
4 \(x \sec x=\tan y+c\)
5 \(x \tan x=y \tan x+c\)
Explanation:
(A) : Given, \(\frac{d y}{d x}+y \tan x=\sec x\) It is a first-degree linear D.E. of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\theta\) Here, \(\mathrm{P}=\tan \mathrm{x}, \theta=\sec \mathrm{x}\) I.F. \(=\mathrm{e}^{\int \operatorname{pdx}}=\mathrm{e}^{\int \tan x \mathrm{~d} x}=\mathrm{e}^{\log \sec \mathrm{x}}=\sec \mathrm{x}\) The solution is given by- \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F} \cdot=\int \mathrm{Q} \cdot \mathrm{I} \cdot \mathrm{F} \cdot \mathrm{dx}+\mathrm{c}\) \(y \cdot \sec x=\int \sec ^{2} x d x+c\) \(\mathrm{y} \cdot \sec \mathrm{x}=\tan \mathrm{x}+\mathrm{c}\)
Kerala CEE-2004
Differential Equation
87573
Solution of the differential equation \(x=1+x y \frac{d y}{d x}+\frac{(x y)^{2}}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{(x y)^{3}}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is
87571
The solution of the differential equation \(\frac{d y}{d x}=\frac{1}{x+y^{2}}\) is
1 \(y=-x^{2}-2 x-2+c^{x}\)
2 \(y=x^{2}+2 x+2-c e^{x}\)
3 \(x=-y^{2}-2 y+2-c e^{y}\)
4 \(x=-y^{2}-2 y-2+c e^{y}\)
5 \(x=y^{2}+2 y+2-c e^{y}\)
Explanation:
(D) : Given differential equation, \(\frac{d y}{d x}=\frac{1}{x+y^{2}} \Rightarrow \frac{d x}{d y}=\frac{x+y^{2}}{1} \Rightarrow \frac{d x}{d y}-x=y^{2}\) Assume, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{y}^{2}\) I.F, \(=e \int-1 d y=e^{-y}\) \(x e^{-y}=\int e^{-y} y^{2} d y=-e^{-y} y^{2}+\int 2 e^{-y} y d y\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y+\int e^{-y} d y\right]+c\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y-e^{-y}\right]+c\) \(\Rightarrow x e^{-y}=e^{-y}\left(-y^{2}-2 y-2\right)+c\) \(x=-y^{2}-2 y-2+c e^{y}\)
Kerala CEE-2009
Differential Equation
87572
The solution of \(\frac{d y}{d x}+y \tan x=\sec x\) is :
1 \(y \sec x=\tan x+c\)
2 \(y \tan x=\sec x+c\)
3 \(\tan x=y \tan x+c\)
4 \(x \sec x=\tan y+c\)
5 \(x \tan x=y \tan x+c\)
Explanation:
(A) : Given, \(\frac{d y}{d x}+y \tan x=\sec x\) It is a first-degree linear D.E. of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\theta\) Here, \(\mathrm{P}=\tan \mathrm{x}, \theta=\sec \mathrm{x}\) I.F. \(=\mathrm{e}^{\int \operatorname{pdx}}=\mathrm{e}^{\int \tan x \mathrm{~d} x}=\mathrm{e}^{\log \sec \mathrm{x}}=\sec \mathrm{x}\) The solution is given by- \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F} \cdot=\int \mathrm{Q} \cdot \mathrm{I} \cdot \mathrm{F} \cdot \mathrm{dx}+\mathrm{c}\) \(y \cdot \sec x=\int \sec ^{2} x d x+c\) \(\mathrm{y} \cdot \sec \mathrm{x}=\tan \mathrm{x}+\mathrm{c}\)
Kerala CEE-2004
Differential Equation
87573
Solution of the differential equation \(x=1+x y \frac{d y}{d x}+\frac{(x y)^{2}}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{(x y)^{3}}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is
87571
The solution of the differential equation \(\frac{d y}{d x}=\frac{1}{x+y^{2}}\) is
1 \(y=-x^{2}-2 x-2+c^{x}\)
2 \(y=x^{2}+2 x+2-c e^{x}\)
3 \(x=-y^{2}-2 y+2-c e^{y}\)
4 \(x=-y^{2}-2 y-2+c e^{y}\)
5 \(x=y^{2}+2 y+2-c e^{y}\)
Explanation:
(D) : Given differential equation, \(\frac{d y}{d x}=\frac{1}{x+y^{2}} \Rightarrow \frac{d x}{d y}=\frac{x+y^{2}}{1} \Rightarrow \frac{d x}{d y}-x=y^{2}\) Assume, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{y}^{2}\) I.F, \(=e \int-1 d y=e^{-y}\) \(x e^{-y}=\int e^{-y} y^{2} d y=-e^{-y} y^{2}+\int 2 e^{-y} y d y\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y+\int e^{-y} d y\right]+c\) \(=-e^{-y} y^{2}+2\left[-e^{-y} y-e^{-y}\right]+c\) \(\Rightarrow x e^{-y}=e^{-y}\left(-y^{2}-2 y-2\right)+c\) \(x=-y^{2}-2 y-2+c e^{y}\)
Kerala CEE-2009
Differential Equation
87572
The solution of \(\frac{d y}{d x}+y \tan x=\sec x\) is :
1 \(y \sec x=\tan x+c\)
2 \(y \tan x=\sec x+c\)
3 \(\tan x=y \tan x+c\)
4 \(x \sec x=\tan y+c\)
5 \(x \tan x=y \tan x+c\)
Explanation:
(A) : Given, \(\frac{d y}{d x}+y \tan x=\sec x\) It is a first-degree linear D.E. of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\theta\) Here, \(\mathrm{P}=\tan \mathrm{x}, \theta=\sec \mathrm{x}\) I.F. \(=\mathrm{e}^{\int \operatorname{pdx}}=\mathrm{e}^{\int \tan x \mathrm{~d} x}=\mathrm{e}^{\log \sec \mathrm{x}}=\sec \mathrm{x}\) The solution is given by- \(\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F} \cdot=\int \mathrm{Q} \cdot \mathrm{I} \cdot \mathrm{F} \cdot \mathrm{dx}+\mathrm{c}\) \(y \cdot \sec x=\int \sec ^{2} x d x+c\) \(\mathrm{y} \cdot \sec \mathrm{x}=\tan \mathrm{x}+\mathrm{c}\)
Kerala CEE-2004
Differential Equation
87573
Solution of the differential equation \(x=1+x y \frac{d y}{d x}+\frac{(x y)^{2}}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{(x y)^{3}}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is
