87545
Which one of the following is correct solution \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y) ?\)
1 \(\sec x=C-2 \sec y\)
2 \(\sec y=C+2 \cos y\)
3 \(\sec y=C-2 \cos x\)
4 \(\sec x=C-2 \cos y\)
Explanation:
(C) : We have \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)\) \(\frac{\mathrm{dy}}{\mathrm{dx}}(\tan \mathrm{y})=2 \sin \mathrm{x} \cos \mathrm{y}\) \(\frac{\sin y d y}{\cos ^{2} y}=2 \sin x d x\) On integrating both side we get \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin x d x\) Put \(\cos y=t\) \(-\sin y d y=d t\) \(-\int \frac{d t}{t^{2}}=2 \int \sin x d x+C\) \(-\left(\frac{1}{-t}\right)=-2 \cos x+C\) \(\frac{1}{t}=-2 \cos x+C\) \(\frac{1}{\cos y}=C-2 \cos x\) \(\sec y=C-2 \cos x\)
AMU-2014
Differential Equation
87546
If \(x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)}\), then \(|f(x y)|\) is equal to
87547
Find the solution of the following differential equation : \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
1 \(x=\tan (x+y)+\sec (x+y)+c\)
2 \(x=\tan (x+y)-\sec (x+y)+c\)
3 \(x=\tan (x+y)+\sec ^{2}(x+y)+c\)
4 \(x=\tan (x+y)-\sec ^{2}(x+y)+c\)
Explanation:
(B) : Given differential equation- \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Put \(\mathrm{x}+\mathrm{y}=\mathrm{t}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\) Or \(\quad \frac{d t}{d x}-1=\frac{d y}{d x}\) Put the above value in equation (i) we get- \(\frac{\mathrm{dt}}{\mathrm{dx}}-1=\sin \mathrm{t} \text { or } \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\mathrm{dx}\) \(\frac{1-\sin \mathrm{t}}{1-\sin ^{2} \mathrm{t}} \mathrm{dt}=\mathrm{dx} \text { or } \frac{1-\sin \mathrm{t}}{\cos ^{2} \mathrm{t}}\) \(\text { Or } \sec ^{2} \mathrm{tdt}-\sec \mathrm{t} \tan \mathrm{t} \mathrm{dt}=\mathrm{dx}\) \(\text { Now, integrating we get- }\) \(\tan \mathrm{t}-\sec \mathrm{t}=\mathrm{x}+\mathrm{C}\) \(\text { Put, } \mathrm{t}=\mathrm{x}+\mathrm{y}\) \(\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}=\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})+\mathrm{C}\)
AP EAMCET-05.10.2021
Differential Equation
87549
In a triangle \(P Q R, A, B\) and \(C\) are the angles opposite the corresponding sides of lengths \(a, b\) and \(c\) respectively. If the side are \(a=5, b=13\) and \(c=12\) then \(\sin \frac{B}{2}+\cos \frac{B}{2}=\)
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\sqrt{2}\)
4 1
Explanation:
(C) : Given, \(a_{2}=5, b=13, c=12\) \(a^{2}+c^{2}=b^{2},\) \((5)^{2}+(12)^{2}=(13)^{2}\) \(169=169\) \(\therefore \quad \triangle \mathrm{PQR}\) is right angled triangle at \(\mathrm{Q}\) \(\angle \mathrm{B}=90^{\circ} \Rightarrow \angle \frac{\mathrm{B}}{2}=45^{\circ}\) So, \(\quad \sin \frac{B}{2}+\cos \frac{B}{2}\) \(=\sin 45^{\circ}+\cos 45^{\circ}\) \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)
J and K CET-2017
Differential Equation
87550
Let \(y=y(x)\) be the solution of the differential equation. \(x \frac{d y}{d x}+y=x \log _{e} x,(x>1)\). If \(2 y(2)=\) \(\log _{e} 4-1\), then \(y(e)\) is equal to
1 \(-\frac{\mathrm{e}}{2}\)
2 \(-\frac{\mathrm{e}^{2}}{2}\)
3 \(\frac{\mathrm{e}}{4}\)
4 \(\frac{\mathrm{e}^{2}}{4}\)
Explanation:
(C) : Given differential equation is- \(\frac{x d y}{d x}+y=x \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{1}{x} y=\log _{e} x \tag{i}\) So, \(\quad\) IF \(=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}} \mathrm{x}}=\mathrm{x}\) Now, solution of differential equation (i), is- \(y \times x=\int\left(\log _{e} x\right) x d x+C\) \(y x=\frac{x^{2}}{2} \log _{e} x-\int \frac{x^{2}}{2} \times \frac{1}{x} d x+C\) (Using integration by parts) \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}+C \tag{ii}\) Given that \(2 y(2)=\log _{e} 4-1\) On substituting, \(x=2\) in equation (ii) we get- \(2 y(2)=\frac{4}{2} \log _{e} 2-\frac{4}{4}+C \tag{iii}\) [Where, \(y(2)\) represents value of \(x\) at \(x=2\) ] \(2 y(2)=\log _{e} 4-1+C \tag{iv}\) \(\left(\because \mathrm{m} \log \mathrm{a}=\log \mathrm{a}^{\mathrm{m}}\right)\) From equation (iii) and (iv), we get- \(\mathrm{C}=0\) So, required solution is, \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}\) Now, at \(x=e, e y(e)=\frac{e^{2}}{2} \log _{e} e-\frac{e^{2}}{4}\) (Where, \(y(e)\) represents value of \(y\) at \(x=e)\) \(y(\mathrm{e})=\frac{\mathrm{e}}{4} \quad\left(\because \log _{\mathrm{e}} \mathrm{e}=1\right)\)
87545
Which one of the following is correct solution \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y) ?\)
1 \(\sec x=C-2 \sec y\)
2 \(\sec y=C+2 \cos y\)
3 \(\sec y=C-2 \cos x\)
4 \(\sec x=C-2 \cos y\)
Explanation:
(C) : We have \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)\) \(\frac{\mathrm{dy}}{\mathrm{dx}}(\tan \mathrm{y})=2 \sin \mathrm{x} \cos \mathrm{y}\) \(\frac{\sin y d y}{\cos ^{2} y}=2 \sin x d x\) On integrating both side we get \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin x d x\) Put \(\cos y=t\) \(-\sin y d y=d t\) \(-\int \frac{d t}{t^{2}}=2 \int \sin x d x+C\) \(-\left(\frac{1}{-t}\right)=-2 \cos x+C\) \(\frac{1}{t}=-2 \cos x+C\) \(\frac{1}{\cos y}=C-2 \cos x\) \(\sec y=C-2 \cos x\)
AMU-2014
Differential Equation
87546
If \(x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)}\), then \(|f(x y)|\) is equal to
87547
Find the solution of the following differential equation : \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
1 \(x=\tan (x+y)+\sec (x+y)+c\)
2 \(x=\tan (x+y)-\sec (x+y)+c\)
3 \(x=\tan (x+y)+\sec ^{2}(x+y)+c\)
4 \(x=\tan (x+y)-\sec ^{2}(x+y)+c\)
Explanation:
(B) : Given differential equation- \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Put \(\mathrm{x}+\mathrm{y}=\mathrm{t}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\) Or \(\quad \frac{d t}{d x}-1=\frac{d y}{d x}\) Put the above value in equation (i) we get- \(\frac{\mathrm{dt}}{\mathrm{dx}}-1=\sin \mathrm{t} \text { or } \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\mathrm{dx}\) \(\frac{1-\sin \mathrm{t}}{1-\sin ^{2} \mathrm{t}} \mathrm{dt}=\mathrm{dx} \text { or } \frac{1-\sin \mathrm{t}}{\cos ^{2} \mathrm{t}}\) \(\text { Or } \sec ^{2} \mathrm{tdt}-\sec \mathrm{t} \tan \mathrm{t} \mathrm{dt}=\mathrm{dx}\) \(\text { Now, integrating we get- }\) \(\tan \mathrm{t}-\sec \mathrm{t}=\mathrm{x}+\mathrm{C}\) \(\text { Put, } \mathrm{t}=\mathrm{x}+\mathrm{y}\) \(\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}=\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})+\mathrm{C}\)
AP EAMCET-05.10.2021
Differential Equation
87549
In a triangle \(P Q R, A, B\) and \(C\) are the angles opposite the corresponding sides of lengths \(a, b\) and \(c\) respectively. If the side are \(a=5, b=13\) and \(c=12\) then \(\sin \frac{B}{2}+\cos \frac{B}{2}=\)
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\sqrt{2}\)
4 1
Explanation:
(C) : Given, \(a_{2}=5, b=13, c=12\) \(a^{2}+c^{2}=b^{2},\) \((5)^{2}+(12)^{2}=(13)^{2}\) \(169=169\) \(\therefore \quad \triangle \mathrm{PQR}\) is right angled triangle at \(\mathrm{Q}\) \(\angle \mathrm{B}=90^{\circ} \Rightarrow \angle \frac{\mathrm{B}}{2}=45^{\circ}\) So, \(\quad \sin \frac{B}{2}+\cos \frac{B}{2}\) \(=\sin 45^{\circ}+\cos 45^{\circ}\) \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)
J and K CET-2017
Differential Equation
87550
Let \(y=y(x)\) be the solution of the differential equation. \(x \frac{d y}{d x}+y=x \log _{e} x,(x>1)\). If \(2 y(2)=\) \(\log _{e} 4-1\), then \(y(e)\) is equal to
1 \(-\frac{\mathrm{e}}{2}\)
2 \(-\frac{\mathrm{e}^{2}}{2}\)
3 \(\frac{\mathrm{e}}{4}\)
4 \(\frac{\mathrm{e}^{2}}{4}\)
Explanation:
(C) : Given differential equation is- \(\frac{x d y}{d x}+y=x \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{1}{x} y=\log _{e} x \tag{i}\) So, \(\quad\) IF \(=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}} \mathrm{x}}=\mathrm{x}\) Now, solution of differential equation (i), is- \(y \times x=\int\left(\log _{e} x\right) x d x+C\) \(y x=\frac{x^{2}}{2} \log _{e} x-\int \frac{x^{2}}{2} \times \frac{1}{x} d x+C\) (Using integration by parts) \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}+C \tag{ii}\) Given that \(2 y(2)=\log _{e} 4-1\) On substituting, \(x=2\) in equation (ii) we get- \(2 y(2)=\frac{4}{2} \log _{e} 2-\frac{4}{4}+C \tag{iii}\) [Where, \(y(2)\) represents value of \(x\) at \(x=2\) ] \(2 y(2)=\log _{e} 4-1+C \tag{iv}\) \(\left(\because \mathrm{m} \log \mathrm{a}=\log \mathrm{a}^{\mathrm{m}}\right)\) From equation (iii) and (iv), we get- \(\mathrm{C}=0\) So, required solution is, \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}\) Now, at \(x=e, e y(e)=\frac{e^{2}}{2} \log _{e} e-\frac{e^{2}}{4}\) (Where, \(y(e)\) represents value of \(y\) at \(x=e)\) \(y(\mathrm{e})=\frac{\mathrm{e}}{4} \quad\left(\because \log _{\mathrm{e}} \mathrm{e}=1\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Differential Equation
87545
Which one of the following is correct solution \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y) ?\)
1 \(\sec x=C-2 \sec y\)
2 \(\sec y=C+2 \cos y\)
3 \(\sec y=C-2 \cos x\)
4 \(\sec x=C-2 \cos y\)
Explanation:
(C) : We have \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)\) \(\frac{\mathrm{dy}}{\mathrm{dx}}(\tan \mathrm{y})=2 \sin \mathrm{x} \cos \mathrm{y}\) \(\frac{\sin y d y}{\cos ^{2} y}=2 \sin x d x\) On integrating both side we get \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin x d x\) Put \(\cos y=t\) \(-\sin y d y=d t\) \(-\int \frac{d t}{t^{2}}=2 \int \sin x d x+C\) \(-\left(\frac{1}{-t}\right)=-2 \cos x+C\) \(\frac{1}{t}=-2 \cos x+C\) \(\frac{1}{\cos y}=C-2 \cos x\) \(\sec y=C-2 \cos x\)
AMU-2014
Differential Equation
87546
If \(x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)}\), then \(|f(x y)|\) is equal to
87547
Find the solution of the following differential equation : \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
1 \(x=\tan (x+y)+\sec (x+y)+c\)
2 \(x=\tan (x+y)-\sec (x+y)+c\)
3 \(x=\tan (x+y)+\sec ^{2}(x+y)+c\)
4 \(x=\tan (x+y)-\sec ^{2}(x+y)+c\)
Explanation:
(B) : Given differential equation- \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Put \(\mathrm{x}+\mathrm{y}=\mathrm{t}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\) Or \(\quad \frac{d t}{d x}-1=\frac{d y}{d x}\) Put the above value in equation (i) we get- \(\frac{\mathrm{dt}}{\mathrm{dx}}-1=\sin \mathrm{t} \text { or } \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\mathrm{dx}\) \(\frac{1-\sin \mathrm{t}}{1-\sin ^{2} \mathrm{t}} \mathrm{dt}=\mathrm{dx} \text { or } \frac{1-\sin \mathrm{t}}{\cos ^{2} \mathrm{t}}\) \(\text { Or } \sec ^{2} \mathrm{tdt}-\sec \mathrm{t} \tan \mathrm{t} \mathrm{dt}=\mathrm{dx}\) \(\text { Now, integrating we get- }\) \(\tan \mathrm{t}-\sec \mathrm{t}=\mathrm{x}+\mathrm{C}\) \(\text { Put, } \mathrm{t}=\mathrm{x}+\mathrm{y}\) \(\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}=\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})+\mathrm{C}\)
AP EAMCET-05.10.2021
Differential Equation
87549
In a triangle \(P Q R, A, B\) and \(C\) are the angles opposite the corresponding sides of lengths \(a, b\) and \(c\) respectively. If the side are \(a=5, b=13\) and \(c=12\) then \(\sin \frac{B}{2}+\cos \frac{B}{2}=\)
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\sqrt{2}\)
4 1
Explanation:
(C) : Given, \(a_{2}=5, b=13, c=12\) \(a^{2}+c^{2}=b^{2},\) \((5)^{2}+(12)^{2}=(13)^{2}\) \(169=169\) \(\therefore \quad \triangle \mathrm{PQR}\) is right angled triangle at \(\mathrm{Q}\) \(\angle \mathrm{B}=90^{\circ} \Rightarrow \angle \frac{\mathrm{B}}{2}=45^{\circ}\) So, \(\quad \sin \frac{B}{2}+\cos \frac{B}{2}\) \(=\sin 45^{\circ}+\cos 45^{\circ}\) \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)
J and K CET-2017
Differential Equation
87550
Let \(y=y(x)\) be the solution of the differential equation. \(x \frac{d y}{d x}+y=x \log _{e} x,(x>1)\). If \(2 y(2)=\) \(\log _{e} 4-1\), then \(y(e)\) is equal to
1 \(-\frac{\mathrm{e}}{2}\)
2 \(-\frac{\mathrm{e}^{2}}{2}\)
3 \(\frac{\mathrm{e}}{4}\)
4 \(\frac{\mathrm{e}^{2}}{4}\)
Explanation:
(C) : Given differential equation is- \(\frac{x d y}{d x}+y=x \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{1}{x} y=\log _{e} x \tag{i}\) So, \(\quad\) IF \(=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}} \mathrm{x}}=\mathrm{x}\) Now, solution of differential equation (i), is- \(y \times x=\int\left(\log _{e} x\right) x d x+C\) \(y x=\frac{x^{2}}{2} \log _{e} x-\int \frac{x^{2}}{2} \times \frac{1}{x} d x+C\) (Using integration by parts) \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}+C \tag{ii}\) Given that \(2 y(2)=\log _{e} 4-1\) On substituting, \(x=2\) in equation (ii) we get- \(2 y(2)=\frac{4}{2} \log _{e} 2-\frac{4}{4}+C \tag{iii}\) [Where, \(y(2)\) represents value of \(x\) at \(x=2\) ] \(2 y(2)=\log _{e} 4-1+C \tag{iv}\) \(\left(\because \mathrm{m} \log \mathrm{a}=\log \mathrm{a}^{\mathrm{m}}\right)\) From equation (iii) and (iv), we get- \(\mathrm{C}=0\) So, required solution is, \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}\) Now, at \(x=e, e y(e)=\frac{e^{2}}{2} \log _{e} e-\frac{e^{2}}{4}\) (Where, \(y(e)\) represents value of \(y\) at \(x=e)\) \(y(\mathrm{e})=\frac{\mathrm{e}}{4} \quad\left(\because \log _{\mathrm{e}} \mathrm{e}=1\right)\)
87545
Which one of the following is correct solution \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y) ?\)
1 \(\sec x=C-2 \sec y\)
2 \(\sec y=C+2 \cos y\)
3 \(\sec y=C-2 \cos x\)
4 \(\sec x=C-2 \cos y\)
Explanation:
(C) : We have \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)\) \(\frac{\mathrm{dy}}{\mathrm{dx}}(\tan \mathrm{y})=2 \sin \mathrm{x} \cos \mathrm{y}\) \(\frac{\sin y d y}{\cos ^{2} y}=2 \sin x d x\) On integrating both side we get \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin x d x\) Put \(\cos y=t\) \(-\sin y d y=d t\) \(-\int \frac{d t}{t^{2}}=2 \int \sin x d x+C\) \(-\left(\frac{1}{-t}\right)=-2 \cos x+C\) \(\frac{1}{t}=-2 \cos x+C\) \(\frac{1}{\cos y}=C-2 \cos x\) \(\sec y=C-2 \cos x\)
AMU-2014
Differential Equation
87546
If \(x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)}\), then \(|f(x y)|\) is equal to
87547
Find the solution of the following differential equation : \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
1 \(x=\tan (x+y)+\sec (x+y)+c\)
2 \(x=\tan (x+y)-\sec (x+y)+c\)
3 \(x=\tan (x+y)+\sec ^{2}(x+y)+c\)
4 \(x=\tan (x+y)-\sec ^{2}(x+y)+c\)
Explanation:
(B) : Given differential equation- \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Put \(\mathrm{x}+\mathrm{y}=\mathrm{t}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\) Or \(\quad \frac{d t}{d x}-1=\frac{d y}{d x}\) Put the above value in equation (i) we get- \(\frac{\mathrm{dt}}{\mathrm{dx}}-1=\sin \mathrm{t} \text { or } \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\mathrm{dx}\) \(\frac{1-\sin \mathrm{t}}{1-\sin ^{2} \mathrm{t}} \mathrm{dt}=\mathrm{dx} \text { or } \frac{1-\sin \mathrm{t}}{\cos ^{2} \mathrm{t}}\) \(\text { Or } \sec ^{2} \mathrm{tdt}-\sec \mathrm{t} \tan \mathrm{t} \mathrm{dt}=\mathrm{dx}\) \(\text { Now, integrating we get- }\) \(\tan \mathrm{t}-\sec \mathrm{t}=\mathrm{x}+\mathrm{C}\) \(\text { Put, } \mathrm{t}=\mathrm{x}+\mathrm{y}\) \(\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}=\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})+\mathrm{C}\)
AP EAMCET-05.10.2021
Differential Equation
87549
In a triangle \(P Q R, A, B\) and \(C\) are the angles opposite the corresponding sides of lengths \(a, b\) and \(c\) respectively. If the side are \(a=5, b=13\) and \(c=12\) then \(\sin \frac{B}{2}+\cos \frac{B}{2}=\)
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\sqrt{2}\)
4 1
Explanation:
(C) : Given, \(a_{2}=5, b=13, c=12\) \(a^{2}+c^{2}=b^{2},\) \((5)^{2}+(12)^{2}=(13)^{2}\) \(169=169\) \(\therefore \quad \triangle \mathrm{PQR}\) is right angled triangle at \(\mathrm{Q}\) \(\angle \mathrm{B}=90^{\circ} \Rightarrow \angle \frac{\mathrm{B}}{2}=45^{\circ}\) So, \(\quad \sin \frac{B}{2}+\cos \frac{B}{2}\) \(=\sin 45^{\circ}+\cos 45^{\circ}\) \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)
J and K CET-2017
Differential Equation
87550
Let \(y=y(x)\) be the solution of the differential equation. \(x \frac{d y}{d x}+y=x \log _{e} x,(x>1)\). If \(2 y(2)=\) \(\log _{e} 4-1\), then \(y(e)\) is equal to
1 \(-\frac{\mathrm{e}}{2}\)
2 \(-\frac{\mathrm{e}^{2}}{2}\)
3 \(\frac{\mathrm{e}}{4}\)
4 \(\frac{\mathrm{e}^{2}}{4}\)
Explanation:
(C) : Given differential equation is- \(\frac{x d y}{d x}+y=x \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{1}{x} y=\log _{e} x \tag{i}\) So, \(\quad\) IF \(=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}} \mathrm{x}}=\mathrm{x}\) Now, solution of differential equation (i), is- \(y \times x=\int\left(\log _{e} x\right) x d x+C\) \(y x=\frac{x^{2}}{2} \log _{e} x-\int \frac{x^{2}}{2} \times \frac{1}{x} d x+C\) (Using integration by parts) \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}+C \tag{ii}\) Given that \(2 y(2)=\log _{e} 4-1\) On substituting, \(x=2\) in equation (ii) we get- \(2 y(2)=\frac{4}{2} \log _{e} 2-\frac{4}{4}+C \tag{iii}\) [Where, \(y(2)\) represents value of \(x\) at \(x=2\) ] \(2 y(2)=\log _{e} 4-1+C \tag{iv}\) \(\left(\because \mathrm{m} \log \mathrm{a}=\log \mathrm{a}^{\mathrm{m}}\right)\) From equation (iii) and (iv), we get- \(\mathrm{C}=0\) So, required solution is, \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}\) Now, at \(x=e, e y(e)=\frac{e^{2}}{2} \log _{e} e-\frac{e^{2}}{4}\) (Where, \(y(e)\) represents value of \(y\) at \(x=e)\) \(y(\mathrm{e})=\frac{\mathrm{e}}{4} \quad\left(\because \log _{\mathrm{e}} \mathrm{e}=1\right)\)
87545
Which one of the following is correct solution \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y) ?\)
1 \(\sec x=C-2 \sec y\)
2 \(\sec y=C+2 \cos y\)
3 \(\sec y=C-2 \cos x\)
4 \(\sec x=C-2 \cos y\)
Explanation:
(C) : We have \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)\) \(\frac{\mathrm{dy}}{\mathrm{dx}}(\tan \mathrm{y})=2 \sin \mathrm{x} \cos \mathrm{y}\) \(\frac{\sin y d y}{\cos ^{2} y}=2 \sin x d x\) On integrating both side we get \(\int \frac{\sin y}{\cos ^{2} y} d y=\int 2 \sin x d x\) Put \(\cos y=t\) \(-\sin y d y=d t\) \(-\int \frac{d t}{t^{2}}=2 \int \sin x d x+C\) \(-\left(\frac{1}{-t}\right)=-2 \cos x+C\) \(\frac{1}{t}=-2 \cos x+C\) \(\frac{1}{\cos y}=C-2 \cos x\) \(\sec y=C-2 \cos x\)
AMU-2014
Differential Equation
87546
If \(x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)}\), then \(|f(x y)|\) is equal to
87547
Find the solution of the following differential equation : \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
1 \(x=\tan (x+y)+\sec (x+y)+c\)
2 \(x=\tan (x+y)-\sec (x+y)+c\)
3 \(x=\tan (x+y)+\sec ^{2}(x+y)+c\)
4 \(x=\tan (x+y)-\sec ^{2}(x+y)+c\)
Explanation:
(B) : Given differential equation- \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\) \(\frac{d y}{d x}=\sin (x+y) \tag{i}\) Put \(\mathrm{x}+\mathrm{y}=\mathrm{t}\) \(1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\) Or \(\quad \frac{d t}{d x}-1=\frac{d y}{d x}\) Put the above value in equation (i) we get- \(\frac{\mathrm{dt}}{\mathrm{dx}}-1=\sin \mathrm{t} \text { or } \frac{\mathrm{dt}}{1+\sin \mathrm{t}}=\mathrm{dx}\) \(\frac{1-\sin \mathrm{t}}{1-\sin ^{2} \mathrm{t}} \mathrm{dt}=\mathrm{dx} \text { or } \frac{1-\sin \mathrm{t}}{\cos ^{2} \mathrm{t}}\) \(\text { Or } \sec ^{2} \mathrm{tdt}-\sec \mathrm{t} \tan \mathrm{t} \mathrm{dt}=\mathrm{dx}\) \(\text { Now, integrating we get- }\) \(\tan \mathrm{t}-\sec \mathrm{t}=\mathrm{x}+\mathrm{C}\) \(\text { Put, } \mathrm{t}=\mathrm{x}+\mathrm{y}\) \(\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{C}\) \(\mathrm{x}=\tan (\mathrm{x}+\mathrm{y})-\sec (\mathrm{x}+\mathrm{y})+\mathrm{C}\)
AP EAMCET-05.10.2021
Differential Equation
87549
In a triangle \(P Q R, A, B\) and \(C\) are the angles opposite the corresponding sides of lengths \(a, b\) and \(c\) respectively. If the side are \(a=5, b=13\) and \(c=12\) then \(\sin \frac{B}{2}+\cos \frac{B}{2}=\)
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\sqrt{2}\)
4 1
Explanation:
(C) : Given, \(a_{2}=5, b=13, c=12\) \(a^{2}+c^{2}=b^{2},\) \((5)^{2}+(12)^{2}=(13)^{2}\) \(169=169\) \(\therefore \quad \triangle \mathrm{PQR}\) is right angled triangle at \(\mathrm{Q}\) \(\angle \mathrm{B}=90^{\circ} \Rightarrow \angle \frac{\mathrm{B}}{2}=45^{\circ}\) So, \(\quad \sin \frac{B}{2}+\cos \frac{B}{2}\) \(=\sin 45^{\circ}+\cos 45^{\circ}\) \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)
J and K CET-2017
Differential Equation
87550
Let \(y=y(x)\) be the solution of the differential equation. \(x \frac{d y}{d x}+y=x \log _{e} x,(x>1)\). If \(2 y(2)=\) \(\log _{e} 4-1\), then \(y(e)\) is equal to
1 \(-\frac{\mathrm{e}}{2}\)
2 \(-\frac{\mathrm{e}^{2}}{2}\)
3 \(\frac{\mathrm{e}}{4}\)
4 \(\frac{\mathrm{e}^{2}}{4}\)
Explanation:
(C) : Given differential equation is- \(\frac{x d y}{d x}+y=x \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{1}{x} y=\log _{e} x \tag{i}\) So, \(\quad\) IF \(=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}} \mathrm{x}}=\mathrm{x}\) Now, solution of differential equation (i), is- \(y \times x=\int\left(\log _{e} x\right) x d x+C\) \(y x=\frac{x^{2}}{2} \log _{e} x-\int \frac{x^{2}}{2} \times \frac{1}{x} d x+C\) (Using integration by parts) \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}+C \tag{ii}\) Given that \(2 y(2)=\log _{e} 4-1\) On substituting, \(x=2\) in equation (ii) we get- \(2 y(2)=\frac{4}{2} \log _{e} 2-\frac{4}{4}+C \tag{iii}\) [Where, \(y(2)\) represents value of \(x\) at \(x=2\) ] \(2 y(2)=\log _{e} 4-1+C \tag{iv}\) \(\left(\because \mathrm{m} \log \mathrm{a}=\log \mathrm{a}^{\mathrm{m}}\right)\) From equation (iii) and (iv), we get- \(\mathrm{C}=0\) So, required solution is, \(y x=\frac{x^{2}}{2} \log _{e} x-\frac{x^{2}}{4}\) Now, at \(x=e, e y(e)=\frac{e^{2}}{2} \log _{e} e-\frac{e^{2}}{4}\) (Where, \(y(e)\) represents value of \(y\) at \(x=e)\) \(y(\mathrm{e})=\frac{\mathrm{e}}{4} \quad\left(\because \log _{\mathrm{e}} \mathrm{e}=1\right)\)
