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Differential Equation

87171 The general solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x}=3 x\) is

1 \(y=x-\frac{c}{x}\)
2 \(y=x+\frac{c}{x}\)
3 \(y=x^{2}-\frac{c}{x}\)
4 \(y=x^{2}+\frac{c}{x}\)
Karnataka CET-2020
Differential Equation

87173 The general solution of
\(\frac{d y}{d x}=\sqrt{1-x^{2}-y^{2}+x^{2} y^{2}}\) is

1 \(2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+c\)
2 \(\cos ^{-1} y=x \cos ^{-1} x+c\)
3 \(\sin ^{-1} \mathrm{y}=\frac{1}{2} \sin ^{-1} \mathrm{x}+\mathrm{c}\)
4 \(2 \sin ^{-1} y=x \sqrt{1-y^{2}}+c\)
Karnataka CET-2011
Differential Equation

87174 The general solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) is

1 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^{2}}{2}+c\)
2 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{y}}\right)^{2}+\mathrm{c}\)
3 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{2}+c\)
4 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{x}}\right)^{2}+\mathrm{c}\)
MHT CET-2020
Differential Equation

87175 The differential equation of the circles having their centres on the line \(\mathrm{y}=8\) and touching the \(\mathrm{X}\)-axis is

1 \((y-8)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=64\)
2 \((y-8)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
3 \(\mathrm{y}^{2}\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=64\)
4 \((y-8)\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
MHT CET-2020
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Differential Equation

87171 The general solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x}=3 x\) is

1 \(y=x-\frac{c}{x}\)
2 \(y=x+\frac{c}{x}\)
3 \(y=x^{2}-\frac{c}{x}\)
4 \(y=x^{2}+\frac{c}{x}\)
Karnataka CET-2020
Differential Equation

87173 The general solution of
\(\frac{d y}{d x}=\sqrt{1-x^{2}-y^{2}+x^{2} y^{2}}\) is

1 \(2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+c\)
2 \(\cos ^{-1} y=x \cos ^{-1} x+c\)
3 \(\sin ^{-1} \mathrm{y}=\frac{1}{2} \sin ^{-1} \mathrm{x}+\mathrm{c}\)
4 \(2 \sin ^{-1} y=x \sqrt{1-y^{2}}+c\)
Karnataka CET-2011
Differential Equation

87174 The general solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) is

1 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^{2}}{2}+c\)
2 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{y}}\right)^{2}+\mathrm{c}\)
3 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{2}+c\)
4 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{x}}\right)^{2}+\mathrm{c}\)
MHT CET-2020
Differential Equation

87175 The differential equation of the circles having their centres on the line \(\mathrm{y}=8\) and touching the \(\mathrm{X}\)-axis is

1 \((y-8)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=64\)
2 \((y-8)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
3 \(\mathrm{y}^{2}\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=64\)
4 \((y-8)\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
MHT CET-2020
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Differential Equation

87171 The general solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x}=3 x\) is

1 \(y=x-\frac{c}{x}\)
2 \(y=x+\frac{c}{x}\)
3 \(y=x^{2}-\frac{c}{x}\)
4 \(y=x^{2}+\frac{c}{x}\)
Karnataka CET-2020
Differential Equation

87173 The general solution of
\(\frac{d y}{d x}=\sqrt{1-x^{2}-y^{2}+x^{2} y^{2}}\) is

1 \(2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+c\)
2 \(\cos ^{-1} y=x \cos ^{-1} x+c\)
3 \(\sin ^{-1} \mathrm{y}=\frac{1}{2} \sin ^{-1} \mathrm{x}+\mathrm{c}\)
4 \(2 \sin ^{-1} y=x \sqrt{1-y^{2}}+c\)
Karnataka CET-2011
Differential Equation

87174 The general solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) is

1 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^{2}}{2}+c\)
2 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{y}}\right)^{2}+\mathrm{c}\)
3 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{2}+c\)
4 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{x}}\right)^{2}+\mathrm{c}\)
MHT CET-2020
Differential Equation

87175 The differential equation of the circles having their centres on the line \(\mathrm{y}=8\) and touching the \(\mathrm{X}\)-axis is

1 \((y-8)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=64\)
2 \((y-8)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
3 \(\mathrm{y}^{2}\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=64\)
4 \((y-8)\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
MHT CET-2020
NEET DIGITAL LIBRARY
Differential Equation

87171 The general solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x}=3 x\) is

1 \(y=x-\frac{c}{x}\)
2 \(y=x+\frac{c}{x}\)
3 \(y=x^{2}-\frac{c}{x}\)
4 \(y=x^{2}+\frac{c}{x}\)
Karnataka CET-2020
Differential Equation

87173 The general solution of
\(\frac{d y}{d x}=\sqrt{1-x^{2}-y^{2}+x^{2} y^{2}}\) is

1 \(2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+c\)
2 \(\cos ^{-1} y=x \cos ^{-1} x+c\)
3 \(\sin ^{-1} \mathrm{y}=\frac{1}{2} \sin ^{-1} \mathrm{x}+\mathrm{c}\)
4 \(2 \sin ^{-1} y=x \sqrt{1-y^{2}}+c\)
Karnataka CET-2011
Differential Equation

87174 The general solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) is

1 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^{2}}{2}+c\)
2 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{y}}\right)^{2}+\mathrm{c}\)
3 \(x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{2}+c\)
4 \(\mathrm{e}^{\tan ^{-1} \mathrm{y}}=\left(\mathrm{e}^{\tan ^{-1} \mathrm{x}}\right)^{2}+\mathrm{c}\)
MHT CET-2020
Differential Equation

87175 The differential equation of the circles having their centres on the line \(\mathrm{y}=8\) and touching the \(\mathrm{X}\)-axis is

1 \((y-8)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=64\)
2 \((y-8)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
3 \(\mathrm{y}^{2}\left(1+\frac{\mathrm{dy}}{\mathrm{dx}}\right)=64\)
4 \((y-8)\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=64\)
MHT CET-2020
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