(A) : Given that, \(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}=-\mathrm{g}\) Multiply by \(\mathrm{m} / \mathrm{k}\) on both side we get- \(\frac{\mathrm{m}}{\mathrm{k}}\left(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}\right)=-\mathrm{g} \frac{\mathrm{m}}{\mathrm{k}} \Rightarrow \frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v}=-\frac{\mathrm{mg}}{\mathrm{k}}\) \(\frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{d}}{\mathrm{dt}}=-\frac{\mathrm{mg}}{\mathrm{k}}-\mathrm{v} \Rightarrow \int \frac{\mathrm{dv}}{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}=-\int \frac{\mathrm{k}}{\mathrm{m}} \mathrm{dt}\) \(\log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)=-\frac{\mathrm{kt}}{\mathrm{m}}+\log \mathrm{c}\) \(\because \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log \mathrm{x} \Rightarrow \log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)-\log \mathrm{c}=-\frac{\mathrm{kt}}{\mathrm{m}}\) We know that, \(\log \frac{a}{b}=\log a-\log b \Rightarrow\) So, \(\log _{e}\left(\frac{\frac{m g}{k}+v}{c}\right)=-\frac{k t}{m}\) \(\frac{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}{\mathrm{c}}=\mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}} \quad\left(\because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \Rightarrow \mathrm{a}=\mathrm{b}^{\mathrm{x}}\right)\) \(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}=\mathrm{ce} \mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}}\) \(\mathrm{v}=\mathrm{ce}^{-\frac{\mathrm{kt}}{\mathrm{m}}}-\frac{\mathrm{mg}}{\mathrm{k}}\)
VITEEE-2018
Differential Equation
87214
If the I.F. of the differential equation \(\frac{d y}{d x}+5 y=\cos x i s \int e^{A d x}\), then \(A=\)
1 0
2 1
3 3
4 5
Explanation:
(D) : Given that, \(\frac{d y}{d x}+5 y=\cos x\) The integrating factor of the differential equation - \(\frac{d y}{d x}+P y=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}\) Here, \(\mathrm{P}=5\) \(\therefore\) I.F. \(=\mathrm{e}^{\int 5 \mathrm{~d} \mathrm{x}}\) Hence, \(\mathrm{A}=5\)
VITEEE-2018
Differential Equation
87215
The solution of the differential equation \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) is
1 \(y \log |x|=C-\frac{1}{2} \cos x\)
2 \(y \log |x|=C+\frac{1}{2} \cos 2 x\)
3 \(y \log |x|=C-\frac{1}{2} \cos 2 x\)
4 \(x y \log |x|=C-\frac{1}{2} \cos 2 x\)
Explanation:
(C) : Given that the differential equation - \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) Dividing by \(\log \mathrm{x}\) in both the side, we get - \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x} \Rightarrow \frac{d y}{d x}+P y=Q\) Solution of the differential equation is \(\text { y. } e^{\int P d x}=\int Q \cdot e^{\int P d x} d x+C\) \(P=\frac{1}{x \log x}, \quad Q=\frac{\sin 2 x}{\log x}\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{1}{x \log x} d x} \Rightarrow I . F=e^{\log (\log x)}\) \(I . F=\log |x| \quad\left(\because e^{\log x}=x\right)\) So, the solution of given differential equation, \(y \cdot \log |x|=\int \frac{\sin 2 x}{\log x} \cdot \log x d x+C\) \(=\int \sin 2 x+C=-\frac{\cos 2 x}{2}+C\) \(\therefore \mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{\cos 2 \mathrm{x}}{2}\) \(\mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{1}{2} \cos 2 \mathrm{x}\)
(A) : Given that, \(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}=-\mathrm{g}\) Multiply by \(\mathrm{m} / \mathrm{k}\) on both side we get- \(\frac{\mathrm{m}}{\mathrm{k}}\left(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}\right)=-\mathrm{g} \frac{\mathrm{m}}{\mathrm{k}} \Rightarrow \frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v}=-\frac{\mathrm{mg}}{\mathrm{k}}\) \(\frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{d}}{\mathrm{dt}}=-\frac{\mathrm{mg}}{\mathrm{k}}-\mathrm{v} \Rightarrow \int \frac{\mathrm{dv}}{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}=-\int \frac{\mathrm{k}}{\mathrm{m}} \mathrm{dt}\) \(\log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)=-\frac{\mathrm{kt}}{\mathrm{m}}+\log \mathrm{c}\) \(\because \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log \mathrm{x} \Rightarrow \log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)-\log \mathrm{c}=-\frac{\mathrm{kt}}{\mathrm{m}}\) We know that, \(\log \frac{a}{b}=\log a-\log b \Rightarrow\) So, \(\log _{e}\left(\frac{\frac{m g}{k}+v}{c}\right)=-\frac{k t}{m}\) \(\frac{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}{\mathrm{c}}=\mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}} \quad\left(\because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \Rightarrow \mathrm{a}=\mathrm{b}^{\mathrm{x}}\right)\) \(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}=\mathrm{ce} \mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}}\) \(\mathrm{v}=\mathrm{ce}^{-\frac{\mathrm{kt}}{\mathrm{m}}}-\frac{\mathrm{mg}}{\mathrm{k}}\)
VITEEE-2018
Differential Equation
87214
If the I.F. of the differential equation \(\frac{d y}{d x}+5 y=\cos x i s \int e^{A d x}\), then \(A=\)
1 0
2 1
3 3
4 5
Explanation:
(D) : Given that, \(\frac{d y}{d x}+5 y=\cos x\) The integrating factor of the differential equation - \(\frac{d y}{d x}+P y=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}\) Here, \(\mathrm{P}=5\) \(\therefore\) I.F. \(=\mathrm{e}^{\int 5 \mathrm{~d} \mathrm{x}}\) Hence, \(\mathrm{A}=5\)
VITEEE-2018
Differential Equation
87215
The solution of the differential equation \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) is
1 \(y \log |x|=C-\frac{1}{2} \cos x\)
2 \(y \log |x|=C+\frac{1}{2} \cos 2 x\)
3 \(y \log |x|=C-\frac{1}{2} \cos 2 x\)
4 \(x y \log |x|=C-\frac{1}{2} \cos 2 x\)
Explanation:
(C) : Given that the differential equation - \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) Dividing by \(\log \mathrm{x}\) in both the side, we get - \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x} \Rightarrow \frac{d y}{d x}+P y=Q\) Solution of the differential equation is \(\text { y. } e^{\int P d x}=\int Q \cdot e^{\int P d x} d x+C\) \(P=\frac{1}{x \log x}, \quad Q=\frac{\sin 2 x}{\log x}\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{1}{x \log x} d x} \Rightarrow I . F=e^{\log (\log x)}\) \(I . F=\log |x| \quad\left(\because e^{\log x}=x\right)\) So, the solution of given differential equation, \(y \cdot \log |x|=\int \frac{\sin 2 x}{\log x} \cdot \log x d x+C\) \(=\int \sin 2 x+C=-\frac{\cos 2 x}{2}+C\) \(\therefore \mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{\cos 2 \mathrm{x}}{2}\) \(\mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{1}{2} \cos 2 \mathrm{x}\)
(A) : Given that, \(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}=-\mathrm{g}\) Multiply by \(\mathrm{m} / \mathrm{k}\) on both side we get- \(\frac{\mathrm{m}}{\mathrm{k}}\left(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}\right)=-\mathrm{g} \frac{\mathrm{m}}{\mathrm{k}} \Rightarrow \frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v}=-\frac{\mathrm{mg}}{\mathrm{k}}\) \(\frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{d}}{\mathrm{dt}}=-\frac{\mathrm{mg}}{\mathrm{k}}-\mathrm{v} \Rightarrow \int \frac{\mathrm{dv}}{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}=-\int \frac{\mathrm{k}}{\mathrm{m}} \mathrm{dt}\) \(\log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)=-\frac{\mathrm{kt}}{\mathrm{m}}+\log \mathrm{c}\) \(\because \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log \mathrm{x} \Rightarrow \log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)-\log \mathrm{c}=-\frac{\mathrm{kt}}{\mathrm{m}}\) We know that, \(\log \frac{a}{b}=\log a-\log b \Rightarrow\) So, \(\log _{e}\left(\frac{\frac{m g}{k}+v}{c}\right)=-\frac{k t}{m}\) \(\frac{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}{\mathrm{c}}=\mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}} \quad\left(\because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \Rightarrow \mathrm{a}=\mathrm{b}^{\mathrm{x}}\right)\) \(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}=\mathrm{ce} \mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}}\) \(\mathrm{v}=\mathrm{ce}^{-\frac{\mathrm{kt}}{\mathrm{m}}}-\frac{\mathrm{mg}}{\mathrm{k}}\)
VITEEE-2018
Differential Equation
87214
If the I.F. of the differential equation \(\frac{d y}{d x}+5 y=\cos x i s \int e^{A d x}\), then \(A=\)
1 0
2 1
3 3
4 5
Explanation:
(D) : Given that, \(\frac{d y}{d x}+5 y=\cos x\) The integrating factor of the differential equation - \(\frac{d y}{d x}+P y=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}\) Here, \(\mathrm{P}=5\) \(\therefore\) I.F. \(=\mathrm{e}^{\int 5 \mathrm{~d} \mathrm{x}}\) Hence, \(\mathrm{A}=5\)
VITEEE-2018
Differential Equation
87215
The solution of the differential equation \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) is
1 \(y \log |x|=C-\frac{1}{2} \cos x\)
2 \(y \log |x|=C+\frac{1}{2} \cos 2 x\)
3 \(y \log |x|=C-\frac{1}{2} \cos 2 x\)
4 \(x y \log |x|=C-\frac{1}{2} \cos 2 x\)
Explanation:
(C) : Given that the differential equation - \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) Dividing by \(\log \mathrm{x}\) in both the side, we get - \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x} \Rightarrow \frac{d y}{d x}+P y=Q\) Solution of the differential equation is \(\text { y. } e^{\int P d x}=\int Q \cdot e^{\int P d x} d x+C\) \(P=\frac{1}{x \log x}, \quad Q=\frac{\sin 2 x}{\log x}\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{1}{x \log x} d x} \Rightarrow I . F=e^{\log (\log x)}\) \(I . F=\log |x| \quad\left(\because e^{\log x}=x\right)\) So, the solution of given differential equation, \(y \cdot \log |x|=\int \frac{\sin 2 x}{\log x} \cdot \log x d x+C\) \(=\int \sin 2 x+C=-\frac{\cos 2 x}{2}+C\) \(\therefore \mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{\cos 2 \mathrm{x}}{2}\) \(\mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{1}{2} \cos 2 \mathrm{x}\)
(A) : Given that, \(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}=-\mathrm{g}\) Multiply by \(\mathrm{m} / \mathrm{k}\) on both side we get- \(\frac{\mathrm{m}}{\mathrm{k}}\left(\frac{\mathrm{dv}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}\right)=-\mathrm{g} \frac{\mathrm{m}}{\mathrm{k}} \Rightarrow \frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v}=-\frac{\mathrm{mg}}{\mathrm{k}}\) \(\frac{\mathrm{m}}{\mathrm{k}} \frac{\mathrm{d}}{\mathrm{dt}}=-\frac{\mathrm{mg}}{\mathrm{k}}-\mathrm{v} \Rightarrow \int \frac{\mathrm{dv}}{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}=-\int \frac{\mathrm{k}}{\mathrm{m}} \mathrm{dt}\) \(\log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)=-\frac{\mathrm{kt}}{\mathrm{m}}+\log \mathrm{c}\) \(\because \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log \mathrm{x} \Rightarrow \log \left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}\right)-\log \mathrm{c}=-\frac{\mathrm{kt}}{\mathrm{m}}\) We know that, \(\log \frac{a}{b}=\log a-\log b \Rightarrow\) So, \(\log _{e}\left(\frac{\frac{m g}{k}+v}{c}\right)=-\frac{k t}{m}\) \(\frac{\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}}{\mathrm{c}}=\mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}} \quad\left(\because \log _{\mathrm{b}} \mathrm{a}=\mathrm{x} \Rightarrow \mathrm{a}=\mathrm{b}^{\mathrm{x}}\right)\) \(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{v}=\mathrm{ce} \mathrm{e}^{-\frac{\mathrm{kt}}{\mathrm{m}}}\) \(\mathrm{v}=\mathrm{ce}^{-\frac{\mathrm{kt}}{\mathrm{m}}}-\frac{\mathrm{mg}}{\mathrm{k}}\)
VITEEE-2018
Differential Equation
87214
If the I.F. of the differential equation \(\frac{d y}{d x}+5 y=\cos x i s \int e^{A d x}\), then \(A=\)
1 0
2 1
3 3
4 5
Explanation:
(D) : Given that, \(\frac{d y}{d x}+5 y=\cos x\) The integrating factor of the differential equation - \(\frac{d y}{d x}+P y=Q\) I.F. \(=\mathrm{e}^{\int \mathrm{Pdx}}\) Here, \(\mathrm{P}=5\) \(\therefore\) I.F. \(=\mathrm{e}^{\int 5 \mathrm{~d} \mathrm{x}}\) Hence, \(\mathrm{A}=5\)
VITEEE-2018
Differential Equation
87215
The solution of the differential equation \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) is
1 \(y \log |x|=C-\frac{1}{2} \cos x\)
2 \(y \log |x|=C+\frac{1}{2} \cos 2 x\)
3 \(y \log |x|=C-\frac{1}{2} \cos 2 x\)
4 \(x y \log |x|=C-\frac{1}{2} \cos 2 x\)
Explanation:
(C) : Given that the differential equation - \(\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x\) Dividing by \(\log \mathrm{x}\) in both the side, we get - \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x} \Rightarrow \frac{d y}{d x}+P y=Q\) Solution of the differential equation is \(\text { y. } e^{\int P d x}=\int Q \cdot e^{\int P d x} d x+C\) \(P=\frac{1}{x \log x}, \quad Q=\frac{\sin 2 x}{\log x}\) \(I \cdot F=e^{\int P d x}=e^{\int \frac{1}{x \log x} d x} \Rightarrow I . F=e^{\log (\log x)}\) \(I . F=\log |x| \quad\left(\because e^{\log x}=x\right)\) So, the solution of given differential equation, \(y \cdot \log |x|=\int \frac{\sin 2 x}{\log x} \cdot \log x d x+C\) \(=\int \sin 2 x+C=-\frac{\cos 2 x}{2}+C\) \(\therefore \mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{\cos 2 \mathrm{x}}{2}\) \(\mathrm{y} \cdot \log |\mathrm{x}|=\mathrm{C}-\frac{1}{2} \cos 2 \mathrm{x}\)
