87508
The differential equation of all straight lines passing through the point \((1,-1)\) is
1 \(y=(x+1) \frac{d y}{d x}+1\)
2 \(y=(x+1) \frac{d y}{d x}-1\)
3 \(y=(x-1) \frac{d y}{d x}+1\)
4 \(y=(x-1) \frac{d y}{d x}-1\)
Explanation:
(D) : Let, \(\mathrm{m}\) be slope of line passing through \((1,-1)\). Equation of line is \(y+1=m(x-1)\) On differentiating w.r.t.x on both sides in equation (i), we get - \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(\mathrm{m}\). \(\mathrm{y}+1=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)(\mathrm{x}-1)\) \(y=(x-1) \frac{d y}{d x}-1\)
COMEDK-2018
Differential Equation
87509
The function \(y\) specified implicitly by the relation \(\int_{0}^{y} e^{t} d t+\int_{0}^{x} \cos t d t=0\) satisfies the differential equation
(D) : : Given, that function \(\int_{0}^{\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}+\int_{0}^{\mathrm{x}} \cos \mathrm{t} \mathrm{dt}=0\) \(\left[\mathrm{e}^{\mathrm{t}}\right]_{0}^{\mathrm{y}}+[\sin \mathrm{t}]_{0}^{\mathrm{x}}=0\) \(\mathrm{e}^{\mathrm{y}}+\sin \mathrm{x}=0\) On differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\cos \mathrm{x}=0\) Again on differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d y}{d x}-\sin \mathrm{x}=0\) \(\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right)=\sin \mathrm{x}\)
UPSEE-2017
Differential Equation
87511
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=3 x \) is:
1 \( x \)
2 \( \ln x \)
3 0
4 \( \infty \)
Explanation:
(A) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=3 x\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=3 \mathrm{x}\) Now integrating factor, we get- I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
UPSEE-2005
Differential Equation
87512
Which of the following is a homogeneous differential equation?
1 \((4 x+6 y+5) d y-(3 y+2 x+4) d x=0\)
2 \(x y d x-\left(x^{3}+y^{3}\right) d y=0\)
3 \(\left(x^{3}+2 y^{2}\right) d y+2 x y_{1} d y=0\)
4 \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\)
Explanation:
(D) : The function \(\mathrm{f}(\mathrm{x}, \mathrm{y})\) is said to be homogenous function of degree \(n\) If \(\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=\lambda^{\mathrm{n}} \mathrm{F}(\mathrm{x}, \mathrm{y})\) \(\lambda\) is on arbitrary constant the homogeneous equation can be written as in the form of \(\frac{\lambda}{x}\) or \(\frac{x}{y}\) Consider \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\) \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) Let \(f(x, y)=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) \(F(\lambda x, \lambda y)=\frac{-(\lambda y)^{2}}{(\lambda y)^{2}+\lambda x \lambda y-(\lambda x)^{2}}\) \(=\frac{-\lambda^{2} y^{2}}{\lambda^{2}\left(x^{2}+x y-y^{2}\right)}=-\lambda^{0}\left(\frac{y^{2}}{x^{2}+x y-y^{2}}\right)=-\lambda^{0} F(x, y)\) On taking equation (i), \(\left(x^{2}-x y-y^{2}\right) d y=-y^{2} \cdot d x\) \(y^{2} \cdot d x+\left(x^{2}-x y-y^{2}\right) d y=0\) This equation is homogeneous.
JCECE-2018
Differential Equation
87513
Which of the following is a homogeneous differential equation \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x, x \neq 0\) is
1 \(\frac{1}{1+\mathrm{x}^{2}}\)
2 \(\log \left(1+x^{2}\right)\)
3 \(1+x^{2}\)
4 \(-\frac{1}{\mathrm{x}}\)
Explanation:
(C) : Given, \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x\) \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \[ \mathrm{P}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \text { and } \mathrm{Q}=\frac{\cot \mathrm{x}}{1+\mathrm{x}^{2}} \]
Now integrating factor \[ I . F=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}} \]
87508
The differential equation of all straight lines passing through the point \((1,-1)\) is
1 \(y=(x+1) \frac{d y}{d x}+1\)
2 \(y=(x+1) \frac{d y}{d x}-1\)
3 \(y=(x-1) \frac{d y}{d x}+1\)
4 \(y=(x-1) \frac{d y}{d x}-1\)
Explanation:
(D) : Let, \(\mathrm{m}\) be slope of line passing through \((1,-1)\). Equation of line is \(y+1=m(x-1)\) On differentiating w.r.t.x on both sides in equation (i), we get - \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(\mathrm{m}\). \(\mathrm{y}+1=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)(\mathrm{x}-1)\) \(y=(x-1) \frac{d y}{d x}-1\)
COMEDK-2018
Differential Equation
87509
The function \(y\) specified implicitly by the relation \(\int_{0}^{y} e^{t} d t+\int_{0}^{x} \cos t d t=0\) satisfies the differential equation
(D) : : Given, that function \(\int_{0}^{\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}+\int_{0}^{\mathrm{x}} \cos \mathrm{t} \mathrm{dt}=0\) \(\left[\mathrm{e}^{\mathrm{t}}\right]_{0}^{\mathrm{y}}+[\sin \mathrm{t}]_{0}^{\mathrm{x}}=0\) \(\mathrm{e}^{\mathrm{y}}+\sin \mathrm{x}=0\) On differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\cos \mathrm{x}=0\) Again on differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d y}{d x}-\sin \mathrm{x}=0\) \(\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right)=\sin \mathrm{x}\)
UPSEE-2017
Differential Equation
87511
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=3 x \) is:
1 \( x \)
2 \( \ln x \)
3 0
4 \( \infty \)
Explanation:
(A) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=3 x\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=3 \mathrm{x}\) Now integrating factor, we get- I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
UPSEE-2005
Differential Equation
87512
Which of the following is a homogeneous differential equation?
1 \((4 x+6 y+5) d y-(3 y+2 x+4) d x=0\)
2 \(x y d x-\left(x^{3}+y^{3}\right) d y=0\)
3 \(\left(x^{3}+2 y^{2}\right) d y+2 x y_{1} d y=0\)
4 \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\)
Explanation:
(D) : The function \(\mathrm{f}(\mathrm{x}, \mathrm{y})\) is said to be homogenous function of degree \(n\) If \(\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=\lambda^{\mathrm{n}} \mathrm{F}(\mathrm{x}, \mathrm{y})\) \(\lambda\) is on arbitrary constant the homogeneous equation can be written as in the form of \(\frac{\lambda}{x}\) or \(\frac{x}{y}\) Consider \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\) \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) Let \(f(x, y)=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) \(F(\lambda x, \lambda y)=\frac{-(\lambda y)^{2}}{(\lambda y)^{2}+\lambda x \lambda y-(\lambda x)^{2}}\) \(=\frac{-\lambda^{2} y^{2}}{\lambda^{2}\left(x^{2}+x y-y^{2}\right)}=-\lambda^{0}\left(\frac{y^{2}}{x^{2}+x y-y^{2}}\right)=-\lambda^{0} F(x, y)\) On taking equation (i), \(\left(x^{2}-x y-y^{2}\right) d y=-y^{2} \cdot d x\) \(y^{2} \cdot d x+\left(x^{2}-x y-y^{2}\right) d y=0\) This equation is homogeneous.
JCECE-2018
Differential Equation
87513
Which of the following is a homogeneous differential equation \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x, x \neq 0\) is
1 \(\frac{1}{1+\mathrm{x}^{2}}\)
2 \(\log \left(1+x^{2}\right)\)
3 \(1+x^{2}\)
4 \(-\frac{1}{\mathrm{x}}\)
Explanation:
(C) : Given, \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x\) \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \[ \mathrm{P}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \text { and } \mathrm{Q}=\frac{\cot \mathrm{x}}{1+\mathrm{x}^{2}} \]
Now integrating factor \[ I . F=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}} \]
87508
The differential equation of all straight lines passing through the point \((1,-1)\) is
1 \(y=(x+1) \frac{d y}{d x}+1\)
2 \(y=(x+1) \frac{d y}{d x}-1\)
3 \(y=(x-1) \frac{d y}{d x}+1\)
4 \(y=(x-1) \frac{d y}{d x}-1\)
Explanation:
(D) : Let, \(\mathrm{m}\) be slope of line passing through \((1,-1)\). Equation of line is \(y+1=m(x-1)\) On differentiating w.r.t.x on both sides in equation (i), we get - \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(\mathrm{m}\). \(\mathrm{y}+1=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)(\mathrm{x}-1)\) \(y=(x-1) \frac{d y}{d x}-1\)
COMEDK-2018
Differential Equation
87509
The function \(y\) specified implicitly by the relation \(\int_{0}^{y} e^{t} d t+\int_{0}^{x} \cos t d t=0\) satisfies the differential equation
(D) : : Given, that function \(\int_{0}^{\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}+\int_{0}^{\mathrm{x}} \cos \mathrm{t} \mathrm{dt}=0\) \(\left[\mathrm{e}^{\mathrm{t}}\right]_{0}^{\mathrm{y}}+[\sin \mathrm{t}]_{0}^{\mathrm{x}}=0\) \(\mathrm{e}^{\mathrm{y}}+\sin \mathrm{x}=0\) On differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\cos \mathrm{x}=0\) Again on differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d y}{d x}-\sin \mathrm{x}=0\) \(\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right)=\sin \mathrm{x}\)
UPSEE-2017
Differential Equation
87511
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=3 x \) is:
1 \( x \)
2 \( \ln x \)
3 0
4 \( \infty \)
Explanation:
(A) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=3 x\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=3 \mathrm{x}\) Now integrating factor, we get- I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
UPSEE-2005
Differential Equation
87512
Which of the following is a homogeneous differential equation?
1 \((4 x+6 y+5) d y-(3 y+2 x+4) d x=0\)
2 \(x y d x-\left(x^{3}+y^{3}\right) d y=0\)
3 \(\left(x^{3}+2 y^{2}\right) d y+2 x y_{1} d y=0\)
4 \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\)
Explanation:
(D) : The function \(\mathrm{f}(\mathrm{x}, \mathrm{y})\) is said to be homogenous function of degree \(n\) If \(\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=\lambda^{\mathrm{n}} \mathrm{F}(\mathrm{x}, \mathrm{y})\) \(\lambda\) is on arbitrary constant the homogeneous equation can be written as in the form of \(\frac{\lambda}{x}\) or \(\frac{x}{y}\) Consider \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\) \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) Let \(f(x, y)=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) \(F(\lambda x, \lambda y)=\frac{-(\lambda y)^{2}}{(\lambda y)^{2}+\lambda x \lambda y-(\lambda x)^{2}}\) \(=\frac{-\lambda^{2} y^{2}}{\lambda^{2}\left(x^{2}+x y-y^{2}\right)}=-\lambda^{0}\left(\frac{y^{2}}{x^{2}+x y-y^{2}}\right)=-\lambda^{0} F(x, y)\) On taking equation (i), \(\left(x^{2}-x y-y^{2}\right) d y=-y^{2} \cdot d x\) \(y^{2} \cdot d x+\left(x^{2}-x y-y^{2}\right) d y=0\) This equation is homogeneous.
JCECE-2018
Differential Equation
87513
Which of the following is a homogeneous differential equation \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x, x \neq 0\) is
1 \(\frac{1}{1+\mathrm{x}^{2}}\)
2 \(\log \left(1+x^{2}\right)\)
3 \(1+x^{2}\)
4 \(-\frac{1}{\mathrm{x}}\)
Explanation:
(C) : Given, \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x\) \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \[ \mathrm{P}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \text { and } \mathrm{Q}=\frac{\cot \mathrm{x}}{1+\mathrm{x}^{2}} \]
Now integrating factor \[ I . F=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}} \]
87508
The differential equation of all straight lines passing through the point \((1,-1)\) is
1 \(y=(x+1) \frac{d y}{d x}+1\)
2 \(y=(x+1) \frac{d y}{d x}-1\)
3 \(y=(x-1) \frac{d y}{d x}+1\)
4 \(y=(x-1) \frac{d y}{d x}-1\)
Explanation:
(D) : Let, \(\mathrm{m}\) be slope of line passing through \((1,-1)\). Equation of line is \(y+1=m(x-1)\) On differentiating w.r.t.x on both sides in equation (i), we get - \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(\mathrm{m}\). \(\mathrm{y}+1=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)(\mathrm{x}-1)\) \(y=(x-1) \frac{d y}{d x}-1\)
COMEDK-2018
Differential Equation
87509
The function \(y\) specified implicitly by the relation \(\int_{0}^{y} e^{t} d t+\int_{0}^{x} \cos t d t=0\) satisfies the differential equation
(D) : : Given, that function \(\int_{0}^{\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}+\int_{0}^{\mathrm{x}} \cos \mathrm{t} \mathrm{dt}=0\) \(\left[\mathrm{e}^{\mathrm{t}}\right]_{0}^{\mathrm{y}}+[\sin \mathrm{t}]_{0}^{\mathrm{x}}=0\) \(\mathrm{e}^{\mathrm{y}}+\sin \mathrm{x}=0\) On differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\cos \mathrm{x}=0\) Again on differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d y}{d x}-\sin \mathrm{x}=0\) \(\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right)=\sin \mathrm{x}\)
UPSEE-2017
Differential Equation
87511
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=3 x \) is:
1 \( x \)
2 \( \ln x \)
3 0
4 \( \infty \)
Explanation:
(A) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=3 x\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=3 \mathrm{x}\) Now integrating factor, we get- I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
UPSEE-2005
Differential Equation
87512
Which of the following is a homogeneous differential equation?
1 \((4 x+6 y+5) d y-(3 y+2 x+4) d x=0\)
2 \(x y d x-\left(x^{3}+y^{3}\right) d y=0\)
3 \(\left(x^{3}+2 y^{2}\right) d y+2 x y_{1} d y=0\)
4 \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\)
Explanation:
(D) : The function \(\mathrm{f}(\mathrm{x}, \mathrm{y})\) is said to be homogenous function of degree \(n\) If \(\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=\lambda^{\mathrm{n}} \mathrm{F}(\mathrm{x}, \mathrm{y})\) \(\lambda\) is on arbitrary constant the homogeneous equation can be written as in the form of \(\frac{\lambda}{x}\) or \(\frac{x}{y}\) Consider \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\) \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) Let \(f(x, y)=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) \(F(\lambda x, \lambda y)=\frac{-(\lambda y)^{2}}{(\lambda y)^{2}+\lambda x \lambda y-(\lambda x)^{2}}\) \(=\frac{-\lambda^{2} y^{2}}{\lambda^{2}\left(x^{2}+x y-y^{2}\right)}=-\lambda^{0}\left(\frac{y^{2}}{x^{2}+x y-y^{2}}\right)=-\lambda^{0} F(x, y)\) On taking equation (i), \(\left(x^{2}-x y-y^{2}\right) d y=-y^{2} \cdot d x\) \(y^{2} \cdot d x+\left(x^{2}-x y-y^{2}\right) d y=0\) This equation is homogeneous.
JCECE-2018
Differential Equation
87513
Which of the following is a homogeneous differential equation \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x, x \neq 0\) is
1 \(\frac{1}{1+\mathrm{x}^{2}}\)
2 \(\log \left(1+x^{2}\right)\)
3 \(1+x^{2}\)
4 \(-\frac{1}{\mathrm{x}}\)
Explanation:
(C) : Given, \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x\) \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \[ \mathrm{P}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \text { and } \mathrm{Q}=\frac{\cot \mathrm{x}}{1+\mathrm{x}^{2}} \]
Now integrating factor \[ I . F=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}} \]
87508
The differential equation of all straight lines passing through the point \((1,-1)\) is
1 \(y=(x+1) \frac{d y}{d x}+1\)
2 \(y=(x+1) \frac{d y}{d x}-1\)
3 \(y=(x-1) \frac{d y}{d x}+1\)
4 \(y=(x-1) \frac{d y}{d x}-1\)
Explanation:
(D) : Let, \(\mathrm{m}\) be slope of line passing through \((1,-1)\). Equation of line is \(y+1=m(x-1)\) On differentiating w.r.t.x on both sides in equation (i), we get - \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(\mathrm{m}\). \(\mathrm{y}+1=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)(\mathrm{x}-1)\) \(y=(x-1) \frac{d y}{d x}-1\)
COMEDK-2018
Differential Equation
87509
The function \(y\) specified implicitly by the relation \(\int_{0}^{y} e^{t} d t+\int_{0}^{x} \cos t d t=0\) satisfies the differential equation
(D) : : Given, that function \(\int_{0}^{\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}+\int_{0}^{\mathrm{x}} \cos \mathrm{t} \mathrm{dt}=0\) \(\left[\mathrm{e}^{\mathrm{t}}\right]_{0}^{\mathrm{y}}+[\sin \mathrm{t}]_{0}^{\mathrm{x}}=0\) \(\mathrm{e}^{\mathrm{y}}+\sin \mathrm{x}=0\) On differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\cos \mathrm{x}=0\) Again on differentiating both sides w.r.t.x, we get \(\mathrm{e}^{\mathrm{y}} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d y}{d x}-\sin \mathrm{x}=0\) \(\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right)=\sin \mathrm{x}\)
UPSEE-2017
Differential Equation
87511
The integrating factor of the differential equation \( \frac{d y}{d x}+\frac{1}{x} y=3 x \) is:
1 \( x \)
2 \( \ln x \)
3 0
4 \( \infty \)
Explanation:
(A) : Given, \(\frac{d y}{d x}+\frac{1}{x} y=3 x\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=3 \mathrm{x}\) Now integrating factor, we get- I.F. \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}\)
UPSEE-2005
Differential Equation
87512
Which of the following is a homogeneous differential equation?
1 \((4 x+6 y+5) d y-(3 y+2 x+4) d x=0\)
2 \(x y d x-\left(x^{3}+y^{3}\right) d y=0\)
3 \(\left(x^{3}+2 y^{2}\right) d y+2 x y_{1} d y=0\)
4 \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\)
Explanation:
(D) : The function \(\mathrm{f}(\mathrm{x}, \mathrm{y})\) is said to be homogenous function of degree \(n\) If \(\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=\lambda^{\mathrm{n}} \mathrm{F}(\mathrm{x}, \mathrm{y})\) \(\lambda\) is on arbitrary constant the homogeneous equation can be written as in the form of \(\frac{\lambda}{x}\) or \(\frac{x}{y}\) Consider \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0\) \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) Let \(f(x, y)=\frac{-y^{2}}{x^{2}-x y-y^{2}}\) \(F(\lambda x, \lambda y)=\frac{-(\lambda y)^{2}}{(\lambda y)^{2}+\lambda x \lambda y-(\lambda x)^{2}}\) \(=\frac{-\lambda^{2} y^{2}}{\lambda^{2}\left(x^{2}+x y-y^{2}\right)}=-\lambda^{0}\left(\frac{y^{2}}{x^{2}+x y-y^{2}}\right)=-\lambda^{0} F(x, y)\) On taking equation (i), \(\left(x^{2}-x y-y^{2}\right) d y=-y^{2} \cdot d x\) \(y^{2} \cdot d x+\left(x^{2}-x y-y^{2}\right) d y=0\) This equation is homogeneous.
JCECE-2018
Differential Equation
87513
Which of the following is a homogeneous differential equation \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x, x \neq 0\) is
1 \(\frac{1}{1+\mathrm{x}^{2}}\)
2 \(\log \left(1+x^{2}\right)\)
3 \(1+x^{2}\)
4 \(-\frac{1}{\mathrm{x}}\)
Explanation:
(C) : Given, \(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x\) \(\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) \[ \mathrm{P}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \text { and } \mathrm{Q}=\frac{\cot \mathrm{x}}{1+\mathrm{x}^{2}} \]
Now integrating factor \[ I . F=\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}} \]
