87180
The differential equation of all lines making intercept 3 on the \(\mathrm{X}\)-axis is
1 \(x \frac{d y}{d x}=3 y\)
2 \((x-3) \frac{d y}{d x}=y\)
3 \(\frac{d y}{d x}=x-3\)
4 \(\frac{d y}{d x}=x+3\)
Explanation:
(B) : Given, differential equation of straight line. \(y=m x+c\) Making intercept 3 on the \(x\) axis then \(y=0\) \(0=3 \mathrm{~m}+\mathrm{c}\) \(\begin{aligned} \therefore \quad \mathrm{c} =-3 \mathrm{~m} \\ \therefore \quad \mathrm{y} =\mathrm{mx}-3 \mathrm{~m}\end{aligned}\) Differentiating both side w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(c\) and \(m\) in equation (i), we get - \(y=m x+(-3 m) \Rightarrow y=m(x-3)\) \(y=\frac{d y}{d x}(x-3)\)
MHT CET-2019
Differential Equation
87181
The particular solution of the differential equation \(\log \left(\frac{d y}{d x}\right)=x\), when \(x=0, y=1\), is
1 \(y=-e^{x}+2\)
2 \(y=e^{x}+2\)
3 \(y=e^{x}\)
4 \(y=-e^{x}\)
Explanation:
(C) : Given, differential equation, \(\log \left(\frac{d y}{d x}\right)=x\) Taking antilog on both sides, we get- \(\frac{d y}{d x}=e^{x} \Rightarrow \quad d y=e^{x} d x\) Integrating on both sides, we get- \(\int d y=\int e^{x} d x \Rightarrow y=e^{x}+c \tag{i}\) When, \(\mathrm{x}=0\) and \(\mathrm{y}=1\) \(1=\mathrm{e}^{0}+\mathrm{c} \Rightarrow 1=1+\mathrm{c}\) \(\mathrm{c}=0\) Putting the value of \(\mathrm{c}\) in equation (i) Then, \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}\)
MHT CET-2019
Differential Equation
87192
The equation of the curve through the point \((1,0)\) and whose slope is \(\frac{y-1}{x^{2}+x}\) is
1 \((\mathrm{y}-1)(\mathrm{x}+1)+2 \mathrm{x}=0\)
2 \(2 x(y-1)+x+1=0\)
3 \(x(y-1)(x+1)+2=0\)
4 \(y(x+1)-x=0\)
Explanation:
(A) : Given, slope \(=\left(\frac{d y}{d x}\right)=\frac{y-1}{x^{2}+x} \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^{2}+x}\) Integrating on both sides we get, \(\int \frac{d y}{y-1} =\int \frac{d x}{x^{2}+x} \Rightarrow \int \frac{d y}{y-1}=\int \frac{d x}{x(x+1)}\) \(\int \frac{d y}{y-1} =\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x\) \(\log (y-1) =\log x-\log (x+1)+\log c\) \(\log (y-1) =\log \left(\frac{c x}{x+1}\right)\) \(y -1=\frac{c x}{x+1} \tag{i}\) Since, curve passes through point \((1,0)\) \(\begin{aligned} 0-1 =\frac{1 . c}{1+1} \\ \mathrm{c} =-2\end{aligned}\) Put the above value of 'c' in equation (i) we get - \(y-1=\frac{-2 x}{x+1}\) \((y-1)(x+1)=-2 x\) \((y-1)(x+1)+2 x=0\)
SRM JEEE-2012
Differential Equation
87193
The solution of the equation \(\frac{d^{2} y}{d^{2}}=e^{-2 x}\) is
87180
The differential equation of all lines making intercept 3 on the \(\mathrm{X}\)-axis is
1 \(x \frac{d y}{d x}=3 y\)
2 \((x-3) \frac{d y}{d x}=y\)
3 \(\frac{d y}{d x}=x-3\)
4 \(\frac{d y}{d x}=x+3\)
Explanation:
(B) : Given, differential equation of straight line. \(y=m x+c\) Making intercept 3 on the \(x\) axis then \(y=0\) \(0=3 \mathrm{~m}+\mathrm{c}\) \(\begin{aligned} \therefore \quad \mathrm{c} =-3 \mathrm{~m} \\ \therefore \quad \mathrm{y} =\mathrm{mx}-3 \mathrm{~m}\end{aligned}\) Differentiating both side w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(c\) and \(m\) in equation (i), we get - \(y=m x+(-3 m) \Rightarrow y=m(x-3)\) \(y=\frac{d y}{d x}(x-3)\)
MHT CET-2019
Differential Equation
87181
The particular solution of the differential equation \(\log \left(\frac{d y}{d x}\right)=x\), when \(x=0, y=1\), is
1 \(y=-e^{x}+2\)
2 \(y=e^{x}+2\)
3 \(y=e^{x}\)
4 \(y=-e^{x}\)
Explanation:
(C) : Given, differential equation, \(\log \left(\frac{d y}{d x}\right)=x\) Taking antilog on both sides, we get- \(\frac{d y}{d x}=e^{x} \Rightarrow \quad d y=e^{x} d x\) Integrating on both sides, we get- \(\int d y=\int e^{x} d x \Rightarrow y=e^{x}+c \tag{i}\) When, \(\mathrm{x}=0\) and \(\mathrm{y}=1\) \(1=\mathrm{e}^{0}+\mathrm{c} \Rightarrow 1=1+\mathrm{c}\) \(\mathrm{c}=0\) Putting the value of \(\mathrm{c}\) in equation (i) Then, \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}\)
MHT CET-2019
Differential Equation
87192
The equation of the curve through the point \((1,0)\) and whose slope is \(\frac{y-1}{x^{2}+x}\) is
1 \((\mathrm{y}-1)(\mathrm{x}+1)+2 \mathrm{x}=0\)
2 \(2 x(y-1)+x+1=0\)
3 \(x(y-1)(x+1)+2=0\)
4 \(y(x+1)-x=0\)
Explanation:
(A) : Given, slope \(=\left(\frac{d y}{d x}\right)=\frac{y-1}{x^{2}+x} \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^{2}+x}\) Integrating on both sides we get, \(\int \frac{d y}{y-1} =\int \frac{d x}{x^{2}+x} \Rightarrow \int \frac{d y}{y-1}=\int \frac{d x}{x(x+1)}\) \(\int \frac{d y}{y-1} =\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x\) \(\log (y-1) =\log x-\log (x+1)+\log c\) \(\log (y-1) =\log \left(\frac{c x}{x+1}\right)\) \(y -1=\frac{c x}{x+1} \tag{i}\) Since, curve passes through point \((1,0)\) \(\begin{aligned} 0-1 =\frac{1 . c}{1+1} \\ \mathrm{c} =-2\end{aligned}\) Put the above value of 'c' in equation (i) we get - \(y-1=\frac{-2 x}{x+1}\) \((y-1)(x+1)=-2 x\) \((y-1)(x+1)+2 x=0\)
SRM JEEE-2012
Differential Equation
87193
The solution of the equation \(\frac{d^{2} y}{d^{2}}=e^{-2 x}\) is
87180
The differential equation of all lines making intercept 3 on the \(\mathrm{X}\)-axis is
1 \(x \frac{d y}{d x}=3 y\)
2 \((x-3) \frac{d y}{d x}=y\)
3 \(\frac{d y}{d x}=x-3\)
4 \(\frac{d y}{d x}=x+3\)
Explanation:
(B) : Given, differential equation of straight line. \(y=m x+c\) Making intercept 3 on the \(x\) axis then \(y=0\) \(0=3 \mathrm{~m}+\mathrm{c}\) \(\begin{aligned} \therefore \quad \mathrm{c} =-3 \mathrm{~m} \\ \therefore \quad \mathrm{y} =\mathrm{mx}-3 \mathrm{~m}\end{aligned}\) Differentiating both side w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(c\) and \(m\) in equation (i), we get - \(y=m x+(-3 m) \Rightarrow y=m(x-3)\) \(y=\frac{d y}{d x}(x-3)\)
MHT CET-2019
Differential Equation
87181
The particular solution of the differential equation \(\log \left(\frac{d y}{d x}\right)=x\), when \(x=0, y=1\), is
1 \(y=-e^{x}+2\)
2 \(y=e^{x}+2\)
3 \(y=e^{x}\)
4 \(y=-e^{x}\)
Explanation:
(C) : Given, differential equation, \(\log \left(\frac{d y}{d x}\right)=x\) Taking antilog on both sides, we get- \(\frac{d y}{d x}=e^{x} \Rightarrow \quad d y=e^{x} d x\) Integrating on both sides, we get- \(\int d y=\int e^{x} d x \Rightarrow y=e^{x}+c \tag{i}\) When, \(\mathrm{x}=0\) and \(\mathrm{y}=1\) \(1=\mathrm{e}^{0}+\mathrm{c} \Rightarrow 1=1+\mathrm{c}\) \(\mathrm{c}=0\) Putting the value of \(\mathrm{c}\) in equation (i) Then, \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}\)
MHT CET-2019
Differential Equation
87192
The equation of the curve through the point \((1,0)\) and whose slope is \(\frac{y-1}{x^{2}+x}\) is
1 \((\mathrm{y}-1)(\mathrm{x}+1)+2 \mathrm{x}=0\)
2 \(2 x(y-1)+x+1=0\)
3 \(x(y-1)(x+1)+2=0\)
4 \(y(x+1)-x=0\)
Explanation:
(A) : Given, slope \(=\left(\frac{d y}{d x}\right)=\frac{y-1}{x^{2}+x} \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^{2}+x}\) Integrating on both sides we get, \(\int \frac{d y}{y-1} =\int \frac{d x}{x^{2}+x} \Rightarrow \int \frac{d y}{y-1}=\int \frac{d x}{x(x+1)}\) \(\int \frac{d y}{y-1} =\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x\) \(\log (y-1) =\log x-\log (x+1)+\log c\) \(\log (y-1) =\log \left(\frac{c x}{x+1}\right)\) \(y -1=\frac{c x}{x+1} \tag{i}\) Since, curve passes through point \((1,0)\) \(\begin{aligned} 0-1 =\frac{1 . c}{1+1} \\ \mathrm{c} =-2\end{aligned}\) Put the above value of 'c' in equation (i) we get - \(y-1=\frac{-2 x}{x+1}\) \((y-1)(x+1)=-2 x\) \((y-1)(x+1)+2 x=0\)
SRM JEEE-2012
Differential Equation
87193
The solution of the equation \(\frac{d^{2} y}{d^{2}}=e^{-2 x}\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87180
The differential equation of all lines making intercept 3 on the \(\mathrm{X}\)-axis is
1 \(x \frac{d y}{d x}=3 y\)
2 \((x-3) \frac{d y}{d x}=y\)
3 \(\frac{d y}{d x}=x-3\)
4 \(\frac{d y}{d x}=x+3\)
Explanation:
(B) : Given, differential equation of straight line. \(y=m x+c\) Making intercept 3 on the \(x\) axis then \(y=0\) \(0=3 \mathrm{~m}+\mathrm{c}\) \(\begin{aligned} \therefore \quad \mathrm{c} =-3 \mathrm{~m} \\ \therefore \quad \mathrm{y} =\mathrm{mx}-3 \mathrm{~m}\end{aligned}\) Differentiating both side w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(c\) and \(m\) in equation (i), we get - \(y=m x+(-3 m) \Rightarrow y=m(x-3)\) \(y=\frac{d y}{d x}(x-3)\)
MHT CET-2019
Differential Equation
87181
The particular solution of the differential equation \(\log \left(\frac{d y}{d x}\right)=x\), when \(x=0, y=1\), is
1 \(y=-e^{x}+2\)
2 \(y=e^{x}+2\)
3 \(y=e^{x}\)
4 \(y=-e^{x}\)
Explanation:
(C) : Given, differential equation, \(\log \left(\frac{d y}{d x}\right)=x\) Taking antilog on both sides, we get- \(\frac{d y}{d x}=e^{x} \Rightarrow \quad d y=e^{x} d x\) Integrating on both sides, we get- \(\int d y=\int e^{x} d x \Rightarrow y=e^{x}+c \tag{i}\) When, \(\mathrm{x}=0\) and \(\mathrm{y}=1\) \(1=\mathrm{e}^{0}+\mathrm{c} \Rightarrow 1=1+\mathrm{c}\) \(\mathrm{c}=0\) Putting the value of \(\mathrm{c}\) in equation (i) Then, \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}\)
MHT CET-2019
Differential Equation
87192
The equation of the curve through the point \((1,0)\) and whose slope is \(\frac{y-1}{x^{2}+x}\) is
1 \((\mathrm{y}-1)(\mathrm{x}+1)+2 \mathrm{x}=0\)
2 \(2 x(y-1)+x+1=0\)
3 \(x(y-1)(x+1)+2=0\)
4 \(y(x+1)-x=0\)
Explanation:
(A) : Given, slope \(=\left(\frac{d y}{d x}\right)=\frac{y-1}{x^{2}+x} \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^{2}+x}\) Integrating on both sides we get, \(\int \frac{d y}{y-1} =\int \frac{d x}{x^{2}+x} \Rightarrow \int \frac{d y}{y-1}=\int \frac{d x}{x(x+1)}\) \(\int \frac{d y}{y-1} =\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x\) \(\log (y-1) =\log x-\log (x+1)+\log c\) \(\log (y-1) =\log \left(\frac{c x}{x+1}\right)\) \(y -1=\frac{c x}{x+1} \tag{i}\) Since, curve passes through point \((1,0)\) \(\begin{aligned} 0-1 =\frac{1 . c}{1+1} \\ \mathrm{c} =-2\end{aligned}\) Put the above value of 'c' in equation (i) we get - \(y-1=\frac{-2 x}{x+1}\) \((y-1)(x+1)=-2 x\) \((y-1)(x+1)+2 x=0\)
SRM JEEE-2012
Differential Equation
87193
The solution of the equation \(\frac{d^{2} y}{d^{2}}=e^{-2 x}\) is
87180
The differential equation of all lines making intercept 3 on the \(\mathrm{X}\)-axis is
1 \(x \frac{d y}{d x}=3 y\)
2 \((x-3) \frac{d y}{d x}=y\)
3 \(\frac{d y}{d x}=x-3\)
4 \(\frac{d y}{d x}=x+3\)
Explanation:
(B) : Given, differential equation of straight line. \(y=m x+c\) Making intercept 3 on the \(x\) axis then \(y=0\) \(0=3 \mathrm{~m}+\mathrm{c}\) \(\begin{aligned} \therefore \quad \mathrm{c} =-3 \mathrm{~m} \\ \therefore \quad \mathrm{y} =\mathrm{mx}-3 \mathrm{~m}\end{aligned}\) Differentiating both side w.r.t. \(\mathrm{x}\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\) Putting the value of \(c\) and \(m\) in equation (i), we get - \(y=m x+(-3 m) \Rightarrow y=m(x-3)\) \(y=\frac{d y}{d x}(x-3)\)
MHT CET-2019
Differential Equation
87181
The particular solution of the differential equation \(\log \left(\frac{d y}{d x}\right)=x\), when \(x=0, y=1\), is
1 \(y=-e^{x}+2\)
2 \(y=e^{x}+2\)
3 \(y=e^{x}\)
4 \(y=-e^{x}\)
Explanation:
(C) : Given, differential equation, \(\log \left(\frac{d y}{d x}\right)=x\) Taking antilog on both sides, we get- \(\frac{d y}{d x}=e^{x} \Rightarrow \quad d y=e^{x} d x\) Integrating on both sides, we get- \(\int d y=\int e^{x} d x \Rightarrow y=e^{x}+c \tag{i}\) When, \(\mathrm{x}=0\) and \(\mathrm{y}=1\) \(1=\mathrm{e}^{0}+\mathrm{c} \Rightarrow 1=1+\mathrm{c}\) \(\mathrm{c}=0\) Putting the value of \(\mathrm{c}\) in equation (i) Then, \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}\)
MHT CET-2019
Differential Equation
87192
The equation of the curve through the point \((1,0)\) and whose slope is \(\frac{y-1}{x^{2}+x}\) is
1 \((\mathrm{y}-1)(\mathrm{x}+1)+2 \mathrm{x}=0\)
2 \(2 x(y-1)+x+1=0\)
3 \(x(y-1)(x+1)+2=0\)
4 \(y(x+1)-x=0\)
Explanation:
(A) : Given, slope \(=\left(\frac{d y}{d x}\right)=\frac{y-1}{x^{2}+x} \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^{2}+x}\) Integrating on both sides we get, \(\int \frac{d y}{y-1} =\int \frac{d x}{x^{2}+x} \Rightarrow \int \frac{d y}{y-1}=\int \frac{d x}{x(x+1)}\) \(\int \frac{d y}{y-1} =\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x\) \(\log (y-1) =\log x-\log (x+1)+\log c\) \(\log (y-1) =\log \left(\frac{c x}{x+1}\right)\) \(y -1=\frac{c x}{x+1} \tag{i}\) Since, curve passes through point \((1,0)\) \(\begin{aligned} 0-1 =\frac{1 . c}{1+1} \\ \mathrm{c} =-2\end{aligned}\) Put the above value of 'c' in equation (i) we get - \(y-1=\frac{-2 x}{x+1}\) \((y-1)(x+1)=-2 x\) \((y-1)(x+1)+2 x=0\)
SRM JEEE-2012
Differential Equation
87193
The solution of the equation \(\frac{d^{2} y}{d^{2}}=e^{-2 x}\) is