(B) : Given, \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) On integrating both side, we get- \(\frac{d y}{d x}=\frac{e^{-2 x}}{-2}+c\) Again integrating, we get- \(y=\frac{e^{-2 x}}{(-2)(-2)}+c x+d\) \(y=\frac{1}{4} e^{-2 x}+c x+d\)
UPSEE-2007
Differential Equation
87234
The differential equation of system of concentric circles with centre \((1,2)\) is:
1 \((x-2)+(y-1) \frac{d y}{d x}=0\)
2 \((x-1)+(y-2) \frac{d y}{d x}=0\)
3 \((x+1) \frac{d y}{d x}+(y-2)=0\)
4 \((x+2) \frac{d y}{d x}+(y-1)=0\)
Explanation:
(B) : We know that general equation of circle with centre \((\mathrm{h}, \mathrm{k})\) and radius \(\pi\) \((x-h)^{2}+(y-k)^{2}=\pi^{2}\) Then substituting we get- \((x-1)^{2}+(y-2)^{2}=\pi^{2}\) Differentiating the equation w. r. tx, we get - \(2(x-1)+2(y-2) \frac{d y}{d x}=0\) \((x-1)+(y-2) \frac{d y}{d x}=0\)
UPSEE-2006
Differential Equation
87235
The solution of the differential equation \(\sec ^{2} x\) tan \(y d x+\sec ^{2} y \tan x d y=0\) is:
(A) : Given that the differential equation\(\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0\) Separating the variable, we get- \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\) differential we get- Let \(\quad \tan \mathrm{x}=\mathrm{t}\) Integrating \(\sec ^{2} x d x=d t\) \(\int_{\mathrm{t}}^{1} \mathrm{dt}=\log \mathrm{t}\) And, \(\quad \int \frac{\sec ^{2} y}{\tan y}=\log (\tan y)\) \(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y}=0\) \(\log (\tan x)+\log (\tan y)=\log c\) \(\log (\tan x . \tan y)=\log c\) \(\tan y \cdot \tan \mathrm{x}=\mathrm{c}\)
UPSEE-2004
Differential Equation
87236
The integrating factor of the differential equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) is
1 \(2 \tan x\)
2 \(\frac{1}{2} \sec ^{2} \mathrm{x}\)
3 \(\log \sec \mathrm{x}\)
4 \(\sec ^{2} x\)
Explanation:
(D): Given, \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=2 \tan \mathrm{x}\) and \(\mathrm{Q}=\sin \mathrm{x}\) \(\because\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 2 \tan x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log \sec \mathrm{x}}=\mathrm{e}^{\log \sec ^{2} \mathrm{x}}\) I.F \(=\sec ^{2} x\)
(B) : Given, \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) On integrating both side, we get- \(\frac{d y}{d x}=\frac{e^{-2 x}}{-2}+c\) Again integrating, we get- \(y=\frac{e^{-2 x}}{(-2)(-2)}+c x+d\) \(y=\frac{1}{4} e^{-2 x}+c x+d\)
UPSEE-2007
Differential Equation
87234
The differential equation of system of concentric circles with centre \((1,2)\) is:
1 \((x-2)+(y-1) \frac{d y}{d x}=0\)
2 \((x-1)+(y-2) \frac{d y}{d x}=0\)
3 \((x+1) \frac{d y}{d x}+(y-2)=0\)
4 \((x+2) \frac{d y}{d x}+(y-1)=0\)
Explanation:
(B) : We know that general equation of circle with centre \((\mathrm{h}, \mathrm{k})\) and radius \(\pi\) \((x-h)^{2}+(y-k)^{2}=\pi^{2}\) Then substituting we get- \((x-1)^{2}+(y-2)^{2}=\pi^{2}\) Differentiating the equation w. r. tx, we get - \(2(x-1)+2(y-2) \frac{d y}{d x}=0\) \((x-1)+(y-2) \frac{d y}{d x}=0\)
UPSEE-2006
Differential Equation
87235
The solution of the differential equation \(\sec ^{2} x\) tan \(y d x+\sec ^{2} y \tan x d y=0\) is:
(A) : Given that the differential equation\(\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0\) Separating the variable, we get- \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\) differential we get- Let \(\quad \tan \mathrm{x}=\mathrm{t}\) Integrating \(\sec ^{2} x d x=d t\) \(\int_{\mathrm{t}}^{1} \mathrm{dt}=\log \mathrm{t}\) And, \(\quad \int \frac{\sec ^{2} y}{\tan y}=\log (\tan y)\) \(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y}=0\) \(\log (\tan x)+\log (\tan y)=\log c\) \(\log (\tan x . \tan y)=\log c\) \(\tan y \cdot \tan \mathrm{x}=\mathrm{c}\)
UPSEE-2004
Differential Equation
87236
The integrating factor of the differential equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) is
1 \(2 \tan x\)
2 \(\frac{1}{2} \sec ^{2} \mathrm{x}\)
3 \(\log \sec \mathrm{x}\)
4 \(\sec ^{2} x\)
Explanation:
(D): Given, \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=2 \tan \mathrm{x}\) and \(\mathrm{Q}=\sin \mathrm{x}\) \(\because\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 2 \tan x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log \sec \mathrm{x}}=\mathrm{e}^{\log \sec ^{2} \mathrm{x}}\) I.F \(=\sec ^{2} x\)
(B) : Given, \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) On integrating both side, we get- \(\frac{d y}{d x}=\frac{e^{-2 x}}{-2}+c\) Again integrating, we get- \(y=\frac{e^{-2 x}}{(-2)(-2)}+c x+d\) \(y=\frac{1}{4} e^{-2 x}+c x+d\)
UPSEE-2007
Differential Equation
87234
The differential equation of system of concentric circles with centre \((1,2)\) is:
1 \((x-2)+(y-1) \frac{d y}{d x}=0\)
2 \((x-1)+(y-2) \frac{d y}{d x}=0\)
3 \((x+1) \frac{d y}{d x}+(y-2)=0\)
4 \((x+2) \frac{d y}{d x}+(y-1)=0\)
Explanation:
(B) : We know that general equation of circle with centre \((\mathrm{h}, \mathrm{k})\) and radius \(\pi\) \((x-h)^{2}+(y-k)^{2}=\pi^{2}\) Then substituting we get- \((x-1)^{2}+(y-2)^{2}=\pi^{2}\) Differentiating the equation w. r. tx, we get - \(2(x-1)+2(y-2) \frac{d y}{d x}=0\) \((x-1)+(y-2) \frac{d y}{d x}=0\)
UPSEE-2006
Differential Equation
87235
The solution of the differential equation \(\sec ^{2} x\) tan \(y d x+\sec ^{2} y \tan x d y=0\) is:
(A) : Given that the differential equation\(\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0\) Separating the variable, we get- \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\) differential we get- Let \(\quad \tan \mathrm{x}=\mathrm{t}\) Integrating \(\sec ^{2} x d x=d t\) \(\int_{\mathrm{t}}^{1} \mathrm{dt}=\log \mathrm{t}\) And, \(\quad \int \frac{\sec ^{2} y}{\tan y}=\log (\tan y)\) \(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y}=0\) \(\log (\tan x)+\log (\tan y)=\log c\) \(\log (\tan x . \tan y)=\log c\) \(\tan y \cdot \tan \mathrm{x}=\mathrm{c}\)
UPSEE-2004
Differential Equation
87236
The integrating factor of the differential equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) is
1 \(2 \tan x\)
2 \(\frac{1}{2} \sec ^{2} \mathrm{x}\)
3 \(\log \sec \mathrm{x}\)
4 \(\sec ^{2} x\)
Explanation:
(D): Given, \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=2 \tan \mathrm{x}\) and \(\mathrm{Q}=\sin \mathrm{x}\) \(\because\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 2 \tan x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log \sec \mathrm{x}}=\mathrm{e}^{\log \sec ^{2} \mathrm{x}}\) I.F \(=\sec ^{2} x\)
(B) : Given, \(\frac{d^{2} y}{d x^{2}}=e^{-2 x}\) On integrating both side, we get- \(\frac{d y}{d x}=\frac{e^{-2 x}}{-2}+c\) Again integrating, we get- \(y=\frac{e^{-2 x}}{(-2)(-2)}+c x+d\) \(y=\frac{1}{4} e^{-2 x}+c x+d\)
UPSEE-2007
Differential Equation
87234
The differential equation of system of concentric circles with centre \((1,2)\) is:
1 \((x-2)+(y-1) \frac{d y}{d x}=0\)
2 \((x-1)+(y-2) \frac{d y}{d x}=0\)
3 \((x+1) \frac{d y}{d x}+(y-2)=0\)
4 \((x+2) \frac{d y}{d x}+(y-1)=0\)
Explanation:
(B) : We know that general equation of circle with centre \((\mathrm{h}, \mathrm{k})\) and radius \(\pi\) \((x-h)^{2}+(y-k)^{2}=\pi^{2}\) Then substituting we get- \((x-1)^{2}+(y-2)^{2}=\pi^{2}\) Differentiating the equation w. r. tx, we get - \(2(x-1)+2(y-2) \frac{d y}{d x}=0\) \((x-1)+(y-2) \frac{d y}{d x}=0\)
UPSEE-2006
Differential Equation
87235
The solution of the differential equation \(\sec ^{2} x\) tan \(y d x+\sec ^{2} y \tan x d y=0\) is:
(A) : Given that the differential equation\(\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0\) Separating the variable, we get- \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\) differential we get- Let \(\quad \tan \mathrm{x}=\mathrm{t}\) Integrating \(\sec ^{2} x d x=d t\) \(\int_{\mathrm{t}}^{1} \mathrm{dt}=\log \mathrm{t}\) And, \(\quad \int \frac{\sec ^{2} y}{\tan y}=\log (\tan y)\) \(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y}=0\) \(\log (\tan x)+\log (\tan y)=\log c\) \(\log (\tan x . \tan y)=\log c\) \(\tan y \cdot \tan \mathrm{x}=\mathrm{c}\)
UPSEE-2004
Differential Equation
87236
The integrating factor of the differential equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) is
1 \(2 \tan x\)
2 \(\frac{1}{2} \sec ^{2} \mathrm{x}\)
3 \(\log \sec \mathrm{x}\)
4 \(\sec ^{2} x\)
Explanation:
(D): Given, \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=2 \tan \mathrm{x}\) and \(\mathrm{Q}=\sin \mathrm{x}\) \(\because\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 2 \tan x \mathrm{dx}}\) \(=\mathrm{e}^{2 \log \sec \mathrm{x}}=\mathrm{e}^{\log \sec ^{2} \mathrm{x}}\) I.F \(=\sec ^{2} x\)
