(D) : We have differential equation, \(3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x\) It can be written as- \(\frac{d y}{d x}+\frac{1}{3 x \log _{e} x} y=\frac{2}{3 x}\) Which is a linear form \(\frac{d y}{d x}+P y=Q\), where \(P\) and \(Q\) are function of \(\mathrm{x}\) and the integrating factor is given by the following formula \(\mathrm{e}^{\int \mathrm{Pdx}}\) \(\therefore \quad I F=\mathrm{e}^{\int \frac{1}{3 x \log _{c} x}} \mathrm{dx}\) Put, \(\quad \log _{\mathrm{e}} \mathrm{x}=\mathrm{t}\) \(\frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\) \(=\mathrm{e}^{1 / 3 \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{1 / 3 \log t}=\mathrm{e}^{\log \mathrm{t}^{1 / 3}}=\mathrm{t}^{1 / 3}=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{1 / 3}\)
WB JEE-2012
Differential Equation
87320
A solution of the differential equation \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) is :
1 \(y=2 x\)
2 \(y=-2 x\)
3 \(y=2 x-4\)
4 \(y=2 x+4\)
Explanation:
(C) : Given, \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) Let, \(\quad \frac{d y}{d x}=p\) \(p^{2}-p x+y=0\) \(y=x p-p^{2}\) On Differentiating both sides w.r.t \(\mathrm{x}\), we get- \(\frac{d y}{d x}=x \cdot \frac{d p}{d x}+p-2 p \cdot \frac{d p}{d x}\) \(\frac{d y}{d x}=(x-2 p) \frac{d p}{d x}+p\) On putting the value \(\frac{d y}{d x}\) in above equation, \(p=(x-2 p) \frac{d p}{d x}+p \Rightarrow(x-2 p) \frac{d p}{d x}=0\) \(\frac{d p}{d x}=0\) On integrating w.r.t. \(\mathrm{x}\), we get- \(p=c\) Then, \(\quad \frac{d y}{d x}=c \Rightarrow y=x . c-c^{2}\) \(c=2\) Then, \(\quad y=2 x-4\)
AMU-2013
Differential Equation
87321
Solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}\) (' \(a\) ' being a constant) is
1 \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
2 \(x y=a \tan C x, C\) is an arbitrary constant
3 \(\frac{x}{a}=\tan \frac{y}{C}, C\) is an arbitrary constant
4 \(x y=\tan (x+C), C\) is an arbitrary constant
Explanation:
(A) : Given, Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \((x+y)^{2} \frac{d y}{d x}=a^{2}\) \(\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1\) \(\therefore \quad \mathrm{v}^{2}\left(\frac{\mathrm{dv}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\) \(\Rightarrow \quad \frac{d v}{d x}=\frac{v^{2}+a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2}+a^{2}} d v=d x\) On integrating both sides, we get \(\int \frac{v^{2}}{v^{2}+a^{2}} d v=\int d x\) \(\Rightarrow \quad \int 1-\frac{\mathrm{a}^{2}}{\mathrm{v}^{2}+\mathrm{a}^{2}} \mathrm{dv}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{v}-\frac{\mathrm{a}^{2}}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{v}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{x}+\mathrm{y}-\mathrm{a} \tan ^{-1} \frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad y=a \tan ^{-1} \frac{x+y}{a}+C^{\prime}\) \(\frac{y-C^{\prime}}{a}=\tan ^{-1} \frac{x+y}{a}\) \(\Rightarrow \quad\left(\frac{x+y}{a}\right)=\tan \left(\frac{y+C}{a}\right)\). Hence, \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
WB JEE-2017
Differential Equation
87322
If \(y=\mathrm{e}^{\mathrm{msin}^{-1} \mathrm{x}}\) then \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0\), where \(k\) is equal to
(D) : We have differential equation, \(3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x\) It can be written as- \(\frac{d y}{d x}+\frac{1}{3 x \log _{e} x} y=\frac{2}{3 x}\) Which is a linear form \(\frac{d y}{d x}+P y=Q\), where \(P\) and \(Q\) are function of \(\mathrm{x}\) and the integrating factor is given by the following formula \(\mathrm{e}^{\int \mathrm{Pdx}}\) \(\therefore \quad I F=\mathrm{e}^{\int \frac{1}{3 x \log _{c} x}} \mathrm{dx}\) Put, \(\quad \log _{\mathrm{e}} \mathrm{x}=\mathrm{t}\) \(\frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\) \(=\mathrm{e}^{1 / 3 \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{1 / 3 \log t}=\mathrm{e}^{\log \mathrm{t}^{1 / 3}}=\mathrm{t}^{1 / 3}=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{1 / 3}\)
WB JEE-2012
Differential Equation
87320
A solution of the differential equation \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) is :
1 \(y=2 x\)
2 \(y=-2 x\)
3 \(y=2 x-4\)
4 \(y=2 x+4\)
Explanation:
(C) : Given, \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) Let, \(\quad \frac{d y}{d x}=p\) \(p^{2}-p x+y=0\) \(y=x p-p^{2}\) On Differentiating both sides w.r.t \(\mathrm{x}\), we get- \(\frac{d y}{d x}=x \cdot \frac{d p}{d x}+p-2 p \cdot \frac{d p}{d x}\) \(\frac{d y}{d x}=(x-2 p) \frac{d p}{d x}+p\) On putting the value \(\frac{d y}{d x}\) in above equation, \(p=(x-2 p) \frac{d p}{d x}+p \Rightarrow(x-2 p) \frac{d p}{d x}=0\) \(\frac{d p}{d x}=0\) On integrating w.r.t. \(\mathrm{x}\), we get- \(p=c\) Then, \(\quad \frac{d y}{d x}=c \Rightarrow y=x . c-c^{2}\) \(c=2\) Then, \(\quad y=2 x-4\)
AMU-2013
Differential Equation
87321
Solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}\) (' \(a\) ' being a constant) is
1 \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
2 \(x y=a \tan C x, C\) is an arbitrary constant
3 \(\frac{x}{a}=\tan \frac{y}{C}, C\) is an arbitrary constant
4 \(x y=\tan (x+C), C\) is an arbitrary constant
Explanation:
(A) : Given, Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \((x+y)^{2} \frac{d y}{d x}=a^{2}\) \(\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1\) \(\therefore \quad \mathrm{v}^{2}\left(\frac{\mathrm{dv}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\) \(\Rightarrow \quad \frac{d v}{d x}=\frac{v^{2}+a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2}+a^{2}} d v=d x\) On integrating both sides, we get \(\int \frac{v^{2}}{v^{2}+a^{2}} d v=\int d x\) \(\Rightarrow \quad \int 1-\frac{\mathrm{a}^{2}}{\mathrm{v}^{2}+\mathrm{a}^{2}} \mathrm{dv}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{v}-\frac{\mathrm{a}^{2}}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{v}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{x}+\mathrm{y}-\mathrm{a} \tan ^{-1} \frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad y=a \tan ^{-1} \frac{x+y}{a}+C^{\prime}\) \(\frac{y-C^{\prime}}{a}=\tan ^{-1} \frac{x+y}{a}\) \(\Rightarrow \quad\left(\frac{x+y}{a}\right)=\tan \left(\frac{y+C}{a}\right)\). Hence, \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
WB JEE-2017
Differential Equation
87322
If \(y=\mathrm{e}^{\mathrm{msin}^{-1} \mathrm{x}}\) then \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0\), where \(k\) is equal to
(D) : We have differential equation, \(3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x\) It can be written as- \(\frac{d y}{d x}+\frac{1}{3 x \log _{e} x} y=\frac{2}{3 x}\) Which is a linear form \(\frac{d y}{d x}+P y=Q\), where \(P\) and \(Q\) are function of \(\mathrm{x}\) and the integrating factor is given by the following formula \(\mathrm{e}^{\int \mathrm{Pdx}}\) \(\therefore \quad I F=\mathrm{e}^{\int \frac{1}{3 x \log _{c} x}} \mathrm{dx}\) Put, \(\quad \log _{\mathrm{e}} \mathrm{x}=\mathrm{t}\) \(\frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\) \(=\mathrm{e}^{1 / 3 \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{1 / 3 \log t}=\mathrm{e}^{\log \mathrm{t}^{1 / 3}}=\mathrm{t}^{1 / 3}=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{1 / 3}\)
WB JEE-2012
Differential Equation
87320
A solution of the differential equation \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) is :
1 \(y=2 x\)
2 \(y=-2 x\)
3 \(y=2 x-4\)
4 \(y=2 x+4\)
Explanation:
(C) : Given, \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) Let, \(\quad \frac{d y}{d x}=p\) \(p^{2}-p x+y=0\) \(y=x p-p^{2}\) On Differentiating both sides w.r.t \(\mathrm{x}\), we get- \(\frac{d y}{d x}=x \cdot \frac{d p}{d x}+p-2 p \cdot \frac{d p}{d x}\) \(\frac{d y}{d x}=(x-2 p) \frac{d p}{d x}+p\) On putting the value \(\frac{d y}{d x}\) in above equation, \(p=(x-2 p) \frac{d p}{d x}+p \Rightarrow(x-2 p) \frac{d p}{d x}=0\) \(\frac{d p}{d x}=0\) On integrating w.r.t. \(\mathrm{x}\), we get- \(p=c\) Then, \(\quad \frac{d y}{d x}=c \Rightarrow y=x . c-c^{2}\) \(c=2\) Then, \(\quad y=2 x-4\)
AMU-2013
Differential Equation
87321
Solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}\) (' \(a\) ' being a constant) is
1 \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
2 \(x y=a \tan C x, C\) is an arbitrary constant
3 \(\frac{x}{a}=\tan \frac{y}{C}, C\) is an arbitrary constant
4 \(x y=\tan (x+C), C\) is an arbitrary constant
Explanation:
(A) : Given, Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \((x+y)^{2} \frac{d y}{d x}=a^{2}\) \(\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1\) \(\therefore \quad \mathrm{v}^{2}\left(\frac{\mathrm{dv}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\) \(\Rightarrow \quad \frac{d v}{d x}=\frac{v^{2}+a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2}+a^{2}} d v=d x\) On integrating both sides, we get \(\int \frac{v^{2}}{v^{2}+a^{2}} d v=\int d x\) \(\Rightarrow \quad \int 1-\frac{\mathrm{a}^{2}}{\mathrm{v}^{2}+\mathrm{a}^{2}} \mathrm{dv}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{v}-\frac{\mathrm{a}^{2}}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{v}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{x}+\mathrm{y}-\mathrm{a} \tan ^{-1} \frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad y=a \tan ^{-1} \frac{x+y}{a}+C^{\prime}\) \(\frac{y-C^{\prime}}{a}=\tan ^{-1} \frac{x+y}{a}\) \(\Rightarrow \quad\left(\frac{x+y}{a}\right)=\tan \left(\frac{y+C}{a}\right)\). Hence, \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
WB JEE-2017
Differential Equation
87322
If \(y=\mathrm{e}^{\mathrm{msin}^{-1} \mathrm{x}}\) then \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0\), where \(k\) is equal to
(D) : We have differential equation, \(3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x\) It can be written as- \(\frac{d y}{d x}+\frac{1}{3 x \log _{e} x} y=\frac{2}{3 x}\) Which is a linear form \(\frac{d y}{d x}+P y=Q\), where \(P\) and \(Q\) are function of \(\mathrm{x}\) and the integrating factor is given by the following formula \(\mathrm{e}^{\int \mathrm{Pdx}}\) \(\therefore \quad I F=\mathrm{e}^{\int \frac{1}{3 x \log _{c} x}} \mathrm{dx}\) Put, \(\quad \log _{\mathrm{e}} \mathrm{x}=\mathrm{t}\) \(\frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\) \(=\mathrm{e}^{1 / 3 \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{1 / 3 \log t}=\mathrm{e}^{\log \mathrm{t}^{1 / 3}}=\mathrm{t}^{1 / 3}=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{1 / 3}\)
WB JEE-2012
Differential Equation
87320
A solution of the differential equation \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) is :
1 \(y=2 x\)
2 \(y=-2 x\)
3 \(y=2 x-4\)
4 \(y=2 x+4\)
Explanation:
(C) : Given, \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) Let, \(\quad \frac{d y}{d x}=p\) \(p^{2}-p x+y=0\) \(y=x p-p^{2}\) On Differentiating both sides w.r.t \(\mathrm{x}\), we get- \(\frac{d y}{d x}=x \cdot \frac{d p}{d x}+p-2 p \cdot \frac{d p}{d x}\) \(\frac{d y}{d x}=(x-2 p) \frac{d p}{d x}+p\) On putting the value \(\frac{d y}{d x}\) in above equation, \(p=(x-2 p) \frac{d p}{d x}+p \Rightarrow(x-2 p) \frac{d p}{d x}=0\) \(\frac{d p}{d x}=0\) On integrating w.r.t. \(\mathrm{x}\), we get- \(p=c\) Then, \(\quad \frac{d y}{d x}=c \Rightarrow y=x . c-c^{2}\) \(c=2\) Then, \(\quad y=2 x-4\)
AMU-2013
Differential Equation
87321
Solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}\) (' \(a\) ' being a constant) is
1 \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
2 \(x y=a \tan C x, C\) is an arbitrary constant
3 \(\frac{x}{a}=\tan \frac{y}{C}, C\) is an arbitrary constant
4 \(x y=\tan (x+C), C\) is an arbitrary constant
Explanation:
(A) : Given, Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \((x+y)^{2} \frac{d y}{d x}=a^{2}\) \(\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1\) \(\therefore \quad \mathrm{v}^{2}\left(\frac{\mathrm{dv}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\) \(\Rightarrow \quad \frac{d v}{d x}=\frac{v^{2}+a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2}+a^{2}} d v=d x\) On integrating both sides, we get \(\int \frac{v^{2}}{v^{2}+a^{2}} d v=\int d x\) \(\Rightarrow \quad \int 1-\frac{\mathrm{a}^{2}}{\mathrm{v}^{2}+\mathrm{a}^{2}} \mathrm{dv}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{v}-\frac{\mathrm{a}^{2}}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{v}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{x}+\mathrm{y}-\mathrm{a} \tan ^{-1} \frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad y=a \tan ^{-1} \frac{x+y}{a}+C^{\prime}\) \(\frac{y-C^{\prime}}{a}=\tan ^{-1} \frac{x+y}{a}\) \(\Rightarrow \quad\left(\frac{x+y}{a}\right)=\tan \left(\frac{y+C}{a}\right)\). Hence, \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
WB JEE-2017
Differential Equation
87322
If \(y=\mathrm{e}^{\mathrm{msin}^{-1} \mathrm{x}}\) then \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0\), where \(k\) is equal to
(D) : We have differential equation, \(3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x\) It can be written as- \(\frac{d y}{d x}+\frac{1}{3 x \log _{e} x} y=\frac{2}{3 x}\) Which is a linear form \(\frac{d y}{d x}+P y=Q\), where \(P\) and \(Q\) are function of \(\mathrm{x}\) and the integrating factor is given by the following formula \(\mathrm{e}^{\int \mathrm{Pdx}}\) \(\therefore \quad I F=\mathrm{e}^{\int \frac{1}{3 x \log _{c} x}} \mathrm{dx}\) Put, \(\quad \log _{\mathrm{e}} \mathrm{x}=\mathrm{t}\) \(\frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\) \(=\mathrm{e}^{1 / 3 \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{1 / 3 \log t}=\mathrm{e}^{\log \mathrm{t}^{1 / 3}}=\mathrm{t}^{1 / 3}=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{1 / 3}\)
WB JEE-2012
Differential Equation
87320
A solution of the differential equation \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) is :
1 \(y=2 x\)
2 \(y=-2 x\)
3 \(y=2 x-4\)
4 \(y=2 x+4\)
Explanation:
(C) : Given, \(\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}+y=0\) Let, \(\quad \frac{d y}{d x}=p\) \(p^{2}-p x+y=0\) \(y=x p-p^{2}\) On Differentiating both sides w.r.t \(\mathrm{x}\), we get- \(\frac{d y}{d x}=x \cdot \frac{d p}{d x}+p-2 p \cdot \frac{d p}{d x}\) \(\frac{d y}{d x}=(x-2 p) \frac{d p}{d x}+p\) On putting the value \(\frac{d y}{d x}\) in above equation, \(p=(x-2 p) \frac{d p}{d x}+p \Rightarrow(x-2 p) \frac{d p}{d x}=0\) \(\frac{d p}{d x}=0\) On integrating w.r.t. \(\mathrm{x}\), we get- \(p=c\) Then, \(\quad \frac{d y}{d x}=c \Rightarrow y=x . c-c^{2}\) \(c=2\) Then, \(\quad y=2 x-4\)
AMU-2013
Differential Equation
87321
Solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}\) (' \(a\) ' being a constant) is
1 \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
2 \(x y=a \tan C x, C\) is an arbitrary constant
3 \(\frac{x}{a}=\tan \frac{y}{C}, C\) is an arbitrary constant
4 \(x y=\tan (x+C), C\) is an arbitrary constant
Explanation:
(A) : Given, Let, \(\quad \mathrm{x}+\mathrm{y}=\mathrm{v}\) \((x+y)^{2} \frac{d y}{d x}=a^{2}\) \(\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1\) \(\therefore \quad \mathrm{v}^{2}\left(\frac{\mathrm{dv}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\) \(\Rightarrow \quad \frac{d v}{d x}=\frac{v^{2}+a^{2}}{v^{2}} \Rightarrow \frac{v^{2}}{v^{2}+a^{2}} d v=d x\) On integrating both sides, we get \(\int \frac{v^{2}}{v^{2}+a^{2}} d v=\int d x\) \(\Rightarrow \quad \int 1-\frac{\mathrm{a}^{2}}{\mathrm{v}^{2}+\mathrm{a}^{2}} \mathrm{dv}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{v}-\frac{\mathrm{a}^{2}}{\mathrm{a}} \tan ^{-1} \frac{\mathrm{v}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad \mathrm{x}+\mathrm{y}-\mathrm{a} \tan ^{-1} \frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}=\mathrm{x}+\mathrm{C}^{\prime}\) \(\Rightarrow \quad y=a \tan ^{-1} \frac{x+y}{a}+C^{\prime}\) \(\frac{y-C^{\prime}}{a}=\tan ^{-1} \frac{x+y}{a}\) \(\Rightarrow \quad\left(\frac{x+y}{a}\right)=\tan \left(\frac{y+C}{a}\right)\). Hence, \(\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C\) is an arbitrary constant
WB JEE-2017
Differential Equation
87322
If \(y=\mathrm{e}^{\mathrm{msin}^{-1} \mathrm{x}}\) then \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0\), where \(k\) is equal to
