87481
If the curve \(y=y(x)\) is the solution of the differential equation \(2\left(x^{2}+x^{5 / 4}\right) d y-y(x+\) \(\left.x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) which passes through the point \(\left(1,1-\frac{4}{3} \log _{e} 2\right)\), then the value of \(\mathbf{y}(16)\) is equal to
(C) : Given, The differential equation, \(2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) After rearranging \(\frac{d y}{d x}-\frac{y}{2 x} =\frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)}=e^{\int-\frac{1}{2 x} d x}=e^{-\frac{1}{2} \int \frac{1}{x} d x}\) \(=e^{-\frac{1}{2} \log x}=e^{\log (x)^{-1 / 2}}=\frac{1}{x^{1 / 2}}\) Its solution is, \(\mathrm{y} \times(\mathrm{IF})=\int \mathrm{Q} \times(\mathrm{IF}) \mathrm{dx}\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)} \times x^{-\frac{1}{2}} d x=\int \frac{(x)^{\frac{9}{4}-\frac{5}{4}-\frac{1}{2}}}{\left(x^{3 / 4}+1\right)} d x\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{1 / 2}}{\left(x^{3 / 4}+1\right)} d x\) On putting \(\mathrm{x}=\mathrm{z}^{4}\) \(d x=4 x^{3} \cdot d z\) RHS of equation (i) becomes, \(\int \frac{z^{2} \cdot 4 z^{3}}{\left(z^{3}+1\right)} d z=4 \int \frac{z^{2}\left(z^{3}+1-1\right)}{\left(z^{3}+1\right)} d z\) \(=4\left[\int \frac{z^{2}\left(z^{3}+1\right)}{\left(z^{3}+1\right)} d z-\int \frac{z^{2}}{\left(z^{3}+1\right)} d z\right]\) \(=4\left[\frac{z^{3}}{3}-\frac{1}{3}, \int \frac{3 z^{2}}{z^{3}+1} d z\right]=4\left[\frac{z^{3}}{3}-\frac{1}{3} \cdot \log \left|z^{3}+1\right|\right]\) \(=\frac{4 z^{3}}{3}-\frac{4}{3} \log \left|z^{3}+1\right|+C \tag{i}\) Putting the value \(x=z^{4}\) and \(x^{3 / 4}=z^{3}\) in equation (i) \(=\frac{4 x^{3 / 4}}{3}-\frac{4}{3} \log \left|x^{3 / 4}+1\right|+C\) \(\mathrm{y} \times \mathrm{x}^{-1 / 2}=\frac{4 \mathrm{x}^{3 / 4}}{3}-\frac{4}{3} \log \left|\mathrm{x}^{3 / 4}+1\right|+\mathrm{C}\) Since, this passes through \(\left(1,1-\frac{4}{3} \log _{\mathrm{e}} 2\right)\) Then, \(\quad\left(1-\frac{4}{3} \log _{\mathrm{e}} 2\right) \times 1=\frac{4 \times 1}{3}-\frac{4}{3} \log |1+1|+\mathrm{C}\) \(\Rightarrow \quad C=1-\frac{4}{3}=-\frac{1}{3}\) Hence, \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left|\mathrm{x}^{3 / 4}+1\right|-\frac{\sqrt{\mathrm{x}}}{3}\) \(\because \quad \mathrm{x}>0\) \(\therefore \quad \mathrm{x}^{3 / 4}>0\) \(\left|\mathrm{x}^{3 / 4}+1\right|=\mathrm{x}^{3 / 4}+1\) \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left(\mathrm{x}^{3 / 4}+1\right)-\frac{\sqrt{\mathrm{x}}}{3}\) \(\text { Putting the value } \mathrm{x}=16\) Putting the value \(x=16\) \(y(16)=\frac{4}{3} \times 32-\frac{4}{3} \times 4 \log 9-\frac{4}{3}=4\left(\frac{31}{3}-\frac{8}{3} \log _{\mathrm{e}} 3\right)\)
JEE Main 17.03.2021
Differential Equation
87482
If a curve \(y=f(x)\), passing through the point (1, 2 ), is the solution of the differential equation, \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\), then \(f\left(\frac{1}{2}\right)\) is equal to
1 \(\frac{1}{1+\log _{\mathrm{e}} 2}\)
2 \(\frac{1}{1-\log _{\mathrm{e}} 2}\)
3 \(1+\log _{\mathrm{e}} 2\)
4 \(\frac{-1}{1+\log _{\mathrm{e}} 2}\)
Explanation:
(A): Given, Differential equation- \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\) \(\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(\therefore\) The differential equation becomes, \(v+x \frac{d v}{d x}=\frac{2 v x^{2}+v^{2} x^{2}}{2 x^{2}}\) \(x \frac{d v}{d x}=\frac{2 v+v^{2}-2 v}{2}\) \(x \frac{d v}{d x}=\frac{v^{2}}{2}\) \(\frac{2 d v}{v^{2}}=\frac{d x}{x}\) \(2\left(-\frac{1}{v}\right)=\log _{e} x+c\) \(\log _{e} x+\frac{2 x}{y}+c=0 \tag{i}\) \(\because\) The curve (i) passes through the point \((1,2)\). So, \(\quad \mathrm{C}=-1\) \(\therefore \quad \log _{\mathrm{e}} \mathrm{x}+\frac{2 \mathrm{x}}{\mathrm{y}}=1\) Now, at \(x=\frac{1}{2}\) \(\log _{e}\left(\frac{1}{2}\right)+\frac{1}{y}=1\) \(\frac{1}{y}=1+\log _{e} 2\) \(y=f(1 / 2)=\frac{1}{1+\log _{e} 2}\)
JEE Main 02.09.2020
Differential Equation
87483
The solution curve of the differential equation, \(\left(1+e^{-x}\right)\left(1+y^{2}\right) \frac{d y}{d x}=y^{2}\), which passes through the point \((0,1)\) is
87481
If the curve \(y=y(x)\) is the solution of the differential equation \(2\left(x^{2}+x^{5 / 4}\right) d y-y(x+\) \(\left.x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) which passes through the point \(\left(1,1-\frac{4}{3} \log _{e} 2\right)\), then the value of \(\mathbf{y}(16)\) is equal to
(C) : Given, The differential equation, \(2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) After rearranging \(\frac{d y}{d x}-\frac{y}{2 x} =\frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)}=e^{\int-\frac{1}{2 x} d x}=e^{-\frac{1}{2} \int \frac{1}{x} d x}\) \(=e^{-\frac{1}{2} \log x}=e^{\log (x)^{-1 / 2}}=\frac{1}{x^{1 / 2}}\) Its solution is, \(\mathrm{y} \times(\mathrm{IF})=\int \mathrm{Q} \times(\mathrm{IF}) \mathrm{dx}\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)} \times x^{-\frac{1}{2}} d x=\int \frac{(x)^{\frac{9}{4}-\frac{5}{4}-\frac{1}{2}}}{\left(x^{3 / 4}+1\right)} d x\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{1 / 2}}{\left(x^{3 / 4}+1\right)} d x\) On putting \(\mathrm{x}=\mathrm{z}^{4}\) \(d x=4 x^{3} \cdot d z\) RHS of equation (i) becomes, \(\int \frac{z^{2} \cdot 4 z^{3}}{\left(z^{3}+1\right)} d z=4 \int \frac{z^{2}\left(z^{3}+1-1\right)}{\left(z^{3}+1\right)} d z\) \(=4\left[\int \frac{z^{2}\left(z^{3}+1\right)}{\left(z^{3}+1\right)} d z-\int \frac{z^{2}}{\left(z^{3}+1\right)} d z\right]\) \(=4\left[\frac{z^{3}}{3}-\frac{1}{3}, \int \frac{3 z^{2}}{z^{3}+1} d z\right]=4\left[\frac{z^{3}}{3}-\frac{1}{3} \cdot \log \left|z^{3}+1\right|\right]\) \(=\frac{4 z^{3}}{3}-\frac{4}{3} \log \left|z^{3}+1\right|+C \tag{i}\) Putting the value \(x=z^{4}\) and \(x^{3 / 4}=z^{3}\) in equation (i) \(=\frac{4 x^{3 / 4}}{3}-\frac{4}{3} \log \left|x^{3 / 4}+1\right|+C\) \(\mathrm{y} \times \mathrm{x}^{-1 / 2}=\frac{4 \mathrm{x}^{3 / 4}}{3}-\frac{4}{3} \log \left|\mathrm{x}^{3 / 4}+1\right|+\mathrm{C}\) Since, this passes through \(\left(1,1-\frac{4}{3} \log _{\mathrm{e}} 2\right)\) Then, \(\quad\left(1-\frac{4}{3} \log _{\mathrm{e}} 2\right) \times 1=\frac{4 \times 1}{3}-\frac{4}{3} \log |1+1|+\mathrm{C}\) \(\Rightarrow \quad C=1-\frac{4}{3}=-\frac{1}{3}\) Hence, \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left|\mathrm{x}^{3 / 4}+1\right|-\frac{\sqrt{\mathrm{x}}}{3}\) \(\because \quad \mathrm{x}>0\) \(\therefore \quad \mathrm{x}^{3 / 4}>0\) \(\left|\mathrm{x}^{3 / 4}+1\right|=\mathrm{x}^{3 / 4}+1\) \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left(\mathrm{x}^{3 / 4}+1\right)-\frac{\sqrt{\mathrm{x}}}{3}\) \(\text { Putting the value } \mathrm{x}=16\) Putting the value \(x=16\) \(y(16)=\frac{4}{3} \times 32-\frac{4}{3} \times 4 \log 9-\frac{4}{3}=4\left(\frac{31}{3}-\frac{8}{3} \log _{\mathrm{e}} 3\right)\)
JEE Main 17.03.2021
Differential Equation
87482
If a curve \(y=f(x)\), passing through the point (1, 2 ), is the solution of the differential equation, \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\), then \(f\left(\frac{1}{2}\right)\) is equal to
1 \(\frac{1}{1+\log _{\mathrm{e}} 2}\)
2 \(\frac{1}{1-\log _{\mathrm{e}} 2}\)
3 \(1+\log _{\mathrm{e}} 2\)
4 \(\frac{-1}{1+\log _{\mathrm{e}} 2}\)
Explanation:
(A): Given, Differential equation- \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\) \(\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(\therefore\) The differential equation becomes, \(v+x \frac{d v}{d x}=\frac{2 v x^{2}+v^{2} x^{2}}{2 x^{2}}\) \(x \frac{d v}{d x}=\frac{2 v+v^{2}-2 v}{2}\) \(x \frac{d v}{d x}=\frac{v^{2}}{2}\) \(\frac{2 d v}{v^{2}}=\frac{d x}{x}\) \(2\left(-\frac{1}{v}\right)=\log _{e} x+c\) \(\log _{e} x+\frac{2 x}{y}+c=0 \tag{i}\) \(\because\) The curve (i) passes through the point \((1,2)\). So, \(\quad \mathrm{C}=-1\) \(\therefore \quad \log _{\mathrm{e}} \mathrm{x}+\frac{2 \mathrm{x}}{\mathrm{y}}=1\) Now, at \(x=\frac{1}{2}\) \(\log _{e}\left(\frac{1}{2}\right)+\frac{1}{y}=1\) \(\frac{1}{y}=1+\log _{e} 2\) \(y=f(1 / 2)=\frac{1}{1+\log _{e} 2}\)
JEE Main 02.09.2020
Differential Equation
87483
The solution curve of the differential equation, \(\left(1+e^{-x}\right)\left(1+y^{2}\right) \frac{d y}{d x}=y^{2}\), which passes through the point \((0,1)\) is
87481
If the curve \(y=y(x)\) is the solution of the differential equation \(2\left(x^{2}+x^{5 / 4}\right) d y-y(x+\) \(\left.x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) which passes through the point \(\left(1,1-\frac{4}{3} \log _{e} 2\right)\), then the value of \(\mathbf{y}(16)\) is equal to
(C) : Given, The differential equation, \(2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) After rearranging \(\frac{d y}{d x}-\frac{y}{2 x} =\frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)}=e^{\int-\frac{1}{2 x} d x}=e^{-\frac{1}{2} \int \frac{1}{x} d x}\) \(=e^{-\frac{1}{2} \log x}=e^{\log (x)^{-1 / 2}}=\frac{1}{x^{1 / 2}}\) Its solution is, \(\mathrm{y} \times(\mathrm{IF})=\int \mathrm{Q} \times(\mathrm{IF}) \mathrm{dx}\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)} \times x^{-\frac{1}{2}} d x=\int \frac{(x)^{\frac{9}{4}-\frac{5}{4}-\frac{1}{2}}}{\left(x^{3 / 4}+1\right)} d x\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{1 / 2}}{\left(x^{3 / 4}+1\right)} d x\) On putting \(\mathrm{x}=\mathrm{z}^{4}\) \(d x=4 x^{3} \cdot d z\) RHS of equation (i) becomes, \(\int \frac{z^{2} \cdot 4 z^{3}}{\left(z^{3}+1\right)} d z=4 \int \frac{z^{2}\left(z^{3}+1-1\right)}{\left(z^{3}+1\right)} d z\) \(=4\left[\int \frac{z^{2}\left(z^{3}+1\right)}{\left(z^{3}+1\right)} d z-\int \frac{z^{2}}{\left(z^{3}+1\right)} d z\right]\) \(=4\left[\frac{z^{3}}{3}-\frac{1}{3}, \int \frac{3 z^{2}}{z^{3}+1} d z\right]=4\left[\frac{z^{3}}{3}-\frac{1}{3} \cdot \log \left|z^{3}+1\right|\right]\) \(=\frac{4 z^{3}}{3}-\frac{4}{3} \log \left|z^{3}+1\right|+C \tag{i}\) Putting the value \(x=z^{4}\) and \(x^{3 / 4}=z^{3}\) in equation (i) \(=\frac{4 x^{3 / 4}}{3}-\frac{4}{3} \log \left|x^{3 / 4}+1\right|+C\) \(\mathrm{y} \times \mathrm{x}^{-1 / 2}=\frac{4 \mathrm{x}^{3 / 4}}{3}-\frac{4}{3} \log \left|\mathrm{x}^{3 / 4}+1\right|+\mathrm{C}\) Since, this passes through \(\left(1,1-\frac{4}{3} \log _{\mathrm{e}} 2\right)\) Then, \(\quad\left(1-\frac{4}{3} \log _{\mathrm{e}} 2\right) \times 1=\frac{4 \times 1}{3}-\frac{4}{3} \log |1+1|+\mathrm{C}\) \(\Rightarrow \quad C=1-\frac{4}{3}=-\frac{1}{3}\) Hence, \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left|\mathrm{x}^{3 / 4}+1\right|-\frac{\sqrt{\mathrm{x}}}{3}\) \(\because \quad \mathrm{x}>0\) \(\therefore \quad \mathrm{x}^{3 / 4}>0\) \(\left|\mathrm{x}^{3 / 4}+1\right|=\mathrm{x}^{3 / 4}+1\) \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left(\mathrm{x}^{3 / 4}+1\right)-\frac{\sqrt{\mathrm{x}}}{3}\) \(\text { Putting the value } \mathrm{x}=16\) Putting the value \(x=16\) \(y(16)=\frac{4}{3} \times 32-\frac{4}{3} \times 4 \log 9-\frac{4}{3}=4\left(\frac{31}{3}-\frac{8}{3} \log _{\mathrm{e}} 3\right)\)
JEE Main 17.03.2021
Differential Equation
87482
If a curve \(y=f(x)\), passing through the point (1, 2 ), is the solution of the differential equation, \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\), then \(f\left(\frac{1}{2}\right)\) is equal to
1 \(\frac{1}{1+\log _{\mathrm{e}} 2}\)
2 \(\frac{1}{1-\log _{\mathrm{e}} 2}\)
3 \(1+\log _{\mathrm{e}} 2\)
4 \(\frac{-1}{1+\log _{\mathrm{e}} 2}\)
Explanation:
(A): Given, Differential equation- \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\) \(\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(\therefore\) The differential equation becomes, \(v+x \frac{d v}{d x}=\frac{2 v x^{2}+v^{2} x^{2}}{2 x^{2}}\) \(x \frac{d v}{d x}=\frac{2 v+v^{2}-2 v}{2}\) \(x \frac{d v}{d x}=\frac{v^{2}}{2}\) \(\frac{2 d v}{v^{2}}=\frac{d x}{x}\) \(2\left(-\frac{1}{v}\right)=\log _{e} x+c\) \(\log _{e} x+\frac{2 x}{y}+c=0 \tag{i}\) \(\because\) The curve (i) passes through the point \((1,2)\). So, \(\quad \mathrm{C}=-1\) \(\therefore \quad \log _{\mathrm{e}} \mathrm{x}+\frac{2 \mathrm{x}}{\mathrm{y}}=1\) Now, at \(x=\frac{1}{2}\) \(\log _{e}\left(\frac{1}{2}\right)+\frac{1}{y}=1\) \(\frac{1}{y}=1+\log _{e} 2\) \(y=f(1 / 2)=\frac{1}{1+\log _{e} 2}\)
JEE Main 02.09.2020
Differential Equation
87483
The solution curve of the differential equation, \(\left(1+e^{-x}\right)\left(1+y^{2}\right) \frac{d y}{d x}=y^{2}\), which passes through the point \((0,1)\) is
87481
If the curve \(y=y(x)\) is the solution of the differential equation \(2\left(x^{2}+x^{5 / 4}\right) d y-y(x+\) \(\left.x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) which passes through the point \(\left(1,1-\frac{4}{3} \log _{e} 2\right)\), then the value of \(\mathbf{y}(16)\) is equal to
(C) : Given, The differential equation, \(2\left(x^{2}+x^{5 / 4}\right) d y-y\left(x+x^{1 / 4}\right) d x=2 x^{9 / 4} d x, x>0\) After rearranging \(\frac{d y}{d x}-\frac{y}{2 x} =\frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)}=e^{\int-\frac{1}{2 x} d x}=e^{-\frac{1}{2} \int \frac{1}{x} d x}\) \(=e^{-\frac{1}{2} \log x}=e^{\log (x)^{-1 / 2}}=\frac{1}{x^{1 / 2}}\) Its solution is, \(\mathrm{y} \times(\mathrm{IF})=\int \mathrm{Q} \times(\mathrm{IF}) \mathrm{dx}\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{9 / 4}}{x^{5 / 4}\left(x^{3 / 4}+1\right)} \times x^{-\frac{1}{2}} d x=\int \frac{(x)^{\frac{9}{4}-\frac{5}{4}-\frac{1}{2}}}{\left(x^{3 / 4}+1\right)} d x\) \(y \times \frac{1}{x^{1 / 2}}=\int \frac{x^{1 / 2}}{\left(x^{3 / 4}+1\right)} d x\) On putting \(\mathrm{x}=\mathrm{z}^{4}\) \(d x=4 x^{3} \cdot d z\) RHS of equation (i) becomes, \(\int \frac{z^{2} \cdot 4 z^{3}}{\left(z^{3}+1\right)} d z=4 \int \frac{z^{2}\left(z^{3}+1-1\right)}{\left(z^{3}+1\right)} d z\) \(=4\left[\int \frac{z^{2}\left(z^{3}+1\right)}{\left(z^{3}+1\right)} d z-\int \frac{z^{2}}{\left(z^{3}+1\right)} d z\right]\) \(=4\left[\frac{z^{3}}{3}-\frac{1}{3}, \int \frac{3 z^{2}}{z^{3}+1} d z\right]=4\left[\frac{z^{3}}{3}-\frac{1}{3} \cdot \log \left|z^{3}+1\right|\right]\) \(=\frac{4 z^{3}}{3}-\frac{4}{3} \log \left|z^{3}+1\right|+C \tag{i}\) Putting the value \(x=z^{4}\) and \(x^{3 / 4}=z^{3}\) in equation (i) \(=\frac{4 x^{3 / 4}}{3}-\frac{4}{3} \log \left|x^{3 / 4}+1\right|+C\) \(\mathrm{y} \times \mathrm{x}^{-1 / 2}=\frac{4 \mathrm{x}^{3 / 4}}{3}-\frac{4}{3} \log \left|\mathrm{x}^{3 / 4}+1\right|+\mathrm{C}\) Since, this passes through \(\left(1,1-\frac{4}{3} \log _{\mathrm{e}} 2\right)\) Then, \(\quad\left(1-\frac{4}{3} \log _{\mathrm{e}} 2\right) \times 1=\frac{4 \times 1}{3}-\frac{4}{3} \log |1+1|+\mathrm{C}\) \(\Rightarrow \quad C=1-\frac{4}{3}=-\frac{1}{3}\) Hence, \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left|\mathrm{x}^{3 / 4}+1\right|-\frac{\sqrt{\mathrm{x}}}{3}\) \(\because \quad \mathrm{x}>0\) \(\therefore \quad \mathrm{x}^{3 / 4}>0\) \(\left|\mathrm{x}^{3 / 4}+1\right|=\mathrm{x}^{3 / 4}+1\) \(\mathrm{y}=\frac{4}{3} \mathrm{x}^{5 / 4}-\frac{4}{3} \sqrt{\mathrm{x}} \log \left(\mathrm{x}^{3 / 4}+1\right)-\frac{\sqrt{\mathrm{x}}}{3}\) \(\text { Putting the value } \mathrm{x}=16\) Putting the value \(x=16\) \(y(16)=\frac{4}{3} \times 32-\frac{4}{3} \times 4 \log 9-\frac{4}{3}=4\left(\frac{31}{3}-\frac{8}{3} \log _{\mathrm{e}} 3\right)\)
JEE Main 17.03.2021
Differential Equation
87482
If a curve \(y=f(x)\), passing through the point (1, 2 ), is the solution of the differential equation, \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\), then \(f\left(\frac{1}{2}\right)\) is equal to
1 \(\frac{1}{1+\log _{\mathrm{e}} 2}\)
2 \(\frac{1}{1-\log _{\mathrm{e}} 2}\)
3 \(1+\log _{\mathrm{e}} 2\)
4 \(\frac{-1}{1+\log _{\mathrm{e}} 2}\)
Explanation:
(A): Given, Differential equation- \(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\) \(\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\) \(y=v x\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) \(\therefore\) The differential equation becomes, \(v+x \frac{d v}{d x}=\frac{2 v x^{2}+v^{2} x^{2}}{2 x^{2}}\) \(x \frac{d v}{d x}=\frac{2 v+v^{2}-2 v}{2}\) \(x \frac{d v}{d x}=\frac{v^{2}}{2}\) \(\frac{2 d v}{v^{2}}=\frac{d x}{x}\) \(2\left(-\frac{1}{v}\right)=\log _{e} x+c\) \(\log _{e} x+\frac{2 x}{y}+c=0 \tag{i}\) \(\because\) The curve (i) passes through the point \((1,2)\). So, \(\quad \mathrm{C}=-1\) \(\therefore \quad \log _{\mathrm{e}} \mathrm{x}+\frac{2 \mathrm{x}}{\mathrm{y}}=1\) Now, at \(x=\frac{1}{2}\) \(\log _{e}\left(\frac{1}{2}\right)+\frac{1}{y}=1\) \(\frac{1}{y}=1+\log _{e} 2\) \(y=f(1 / 2)=\frac{1}{1+\log _{e} 2}\)
JEE Main 02.09.2020
Differential Equation
87483
The solution curve of the differential equation, \(\left(1+e^{-x}\right)\left(1+y^{2}\right) \frac{d y}{d x}=y^{2}\), which passes through the point \((0,1)\) is
