(B) : Given differential equation- \(\frac{d y}{d x}+y g^{\prime}(x)=g(x) g^{\prime}(x)\) Which is a linear differential equation- \(\mathrm{P}=\mathrm{g}^{\prime}(\mathrm{x}) \text { and } \mathrm{Q}=\mathrm{g}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})\) \(\begin{array}{ll}\text { I.F, } & =\mathrm{e}^{\int \mathrm{pdx}} \\ \text { I.F, } & =\mathrm{e}^{\mathrm{g}(\mathrm{x})}\end{array}\) For general solution- \(y \cdot e^{g(x)}=\int e^{g(x)} \cdot g(x) g^{\prime}(x) d x+c\). Let, \(g(x)=t \Rightarrow g^{\prime}(x) d x=d t\) \(y \cdot e^{g(x)}=\int e^{t} \cdot t d t \Rightarrow y \cdot e^{g(x)}=t e^{t}-e^{t}+c\) \(y \cdot e^{g(x)}=g(x) e^{g(x)}-e^{g(x)}+c\) \(e^{g(x)}(y-g(x)+1)=c\) Taking \(\log\) on both side- \(\mathrm{g}(\mathrm{x})+\log (1+\mathrm{y}-\mathrm{g}(\mathrm{x}))=\mathrm{c}\)
AP EAMCET-18.09.2020
Differential Equation
87300
If \(y=y(x)\) is the solution of the differential equation \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) with \(y(0)=1\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{4}{3}\)
Explanation:
(A) : We have differential equation- \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}=-\cos x \Rightarrow \frac{d y}{y+1}=-\frac{\cos x}{2+\sin x} d x\) Integrating on both side- Given, \(\int \frac{d y}{y+1}=\int-\frac{\cos x}{2+\sin x} d x\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \((y+1)(2+\sin x)=c\) \(y(0)=1\) \((1+1)(2+0)=c\) \(c=4\) Then equation become- \((y+1)(2+\sin x)=4\) Now, \(\mathrm{y}(\pi / 2)\) we get- \((y+1)(2+\sin \pi / 2)=4 \Rightarrow(y+1)(2+1)=4\) \((y+1)=\frac{4}{3} \Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\)
(B) : Given differential equation- \(\frac{d y}{d x}+y g^{\prime}(x)=g(x) g^{\prime}(x)\) Which is a linear differential equation- \(\mathrm{P}=\mathrm{g}^{\prime}(\mathrm{x}) \text { and } \mathrm{Q}=\mathrm{g}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})\) \(\begin{array}{ll}\text { I.F, } & =\mathrm{e}^{\int \mathrm{pdx}} \\ \text { I.F, } & =\mathrm{e}^{\mathrm{g}(\mathrm{x})}\end{array}\) For general solution- \(y \cdot e^{g(x)}=\int e^{g(x)} \cdot g(x) g^{\prime}(x) d x+c\). Let, \(g(x)=t \Rightarrow g^{\prime}(x) d x=d t\) \(y \cdot e^{g(x)}=\int e^{t} \cdot t d t \Rightarrow y \cdot e^{g(x)}=t e^{t}-e^{t}+c\) \(y \cdot e^{g(x)}=g(x) e^{g(x)}-e^{g(x)}+c\) \(e^{g(x)}(y-g(x)+1)=c\) Taking \(\log\) on both side- \(\mathrm{g}(\mathrm{x})+\log (1+\mathrm{y}-\mathrm{g}(\mathrm{x}))=\mathrm{c}\)
AP EAMCET-18.09.2020
Differential Equation
87300
If \(y=y(x)\) is the solution of the differential equation \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) with \(y(0)=1\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{4}{3}\)
Explanation:
(A) : We have differential equation- \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}=-\cos x \Rightarrow \frac{d y}{y+1}=-\frac{\cos x}{2+\sin x} d x\) Integrating on both side- Given, \(\int \frac{d y}{y+1}=\int-\frac{\cos x}{2+\sin x} d x\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \((y+1)(2+\sin x)=c\) \(y(0)=1\) \((1+1)(2+0)=c\) \(c=4\) Then equation become- \((y+1)(2+\sin x)=4\) Now, \(\mathrm{y}(\pi / 2)\) we get- \((y+1)(2+\sin \pi / 2)=4 \Rightarrow(y+1)(2+1)=4\) \((y+1)=\frac{4}{3} \Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\)
(B) : Given differential equation- \(\frac{d y}{d x}+y g^{\prime}(x)=g(x) g^{\prime}(x)\) Which is a linear differential equation- \(\mathrm{P}=\mathrm{g}^{\prime}(\mathrm{x}) \text { and } \mathrm{Q}=\mathrm{g}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})\) \(\begin{array}{ll}\text { I.F, } & =\mathrm{e}^{\int \mathrm{pdx}} \\ \text { I.F, } & =\mathrm{e}^{\mathrm{g}(\mathrm{x})}\end{array}\) For general solution- \(y \cdot e^{g(x)}=\int e^{g(x)} \cdot g(x) g^{\prime}(x) d x+c\). Let, \(g(x)=t \Rightarrow g^{\prime}(x) d x=d t\) \(y \cdot e^{g(x)}=\int e^{t} \cdot t d t \Rightarrow y \cdot e^{g(x)}=t e^{t}-e^{t}+c\) \(y \cdot e^{g(x)}=g(x) e^{g(x)}-e^{g(x)}+c\) \(e^{g(x)}(y-g(x)+1)=c\) Taking \(\log\) on both side- \(\mathrm{g}(\mathrm{x})+\log (1+\mathrm{y}-\mathrm{g}(\mathrm{x}))=\mathrm{c}\)
AP EAMCET-18.09.2020
Differential Equation
87300
If \(y=y(x)\) is the solution of the differential equation \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) with \(y(0)=1\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{4}{3}\)
Explanation:
(A) : We have differential equation- \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}=-\cos x \Rightarrow \frac{d y}{y+1}=-\frac{\cos x}{2+\sin x} d x\) Integrating on both side- Given, \(\int \frac{d y}{y+1}=\int-\frac{\cos x}{2+\sin x} d x\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \((y+1)(2+\sin x)=c\) \(y(0)=1\) \((1+1)(2+0)=c\) \(c=4\) Then equation become- \((y+1)(2+\sin x)=4\) Now, \(\mathrm{y}(\pi / 2)\) we get- \((y+1)(2+\sin \pi / 2)=4 \Rightarrow(y+1)(2+1)=4\) \((y+1)=\frac{4}{3} \Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\)
(B) : Given differential equation- \(\frac{d y}{d x}+y g^{\prime}(x)=g(x) g^{\prime}(x)\) Which is a linear differential equation- \(\mathrm{P}=\mathrm{g}^{\prime}(\mathrm{x}) \text { and } \mathrm{Q}=\mathrm{g}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})\) \(\begin{array}{ll}\text { I.F, } & =\mathrm{e}^{\int \mathrm{pdx}} \\ \text { I.F, } & =\mathrm{e}^{\mathrm{g}(\mathrm{x})}\end{array}\) For general solution- \(y \cdot e^{g(x)}=\int e^{g(x)} \cdot g(x) g^{\prime}(x) d x+c\). Let, \(g(x)=t \Rightarrow g^{\prime}(x) d x=d t\) \(y \cdot e^{g(x)}=\int e^{t} \cdot t d t \Rightarrow y \cdot e^{g(x)}=t e^{t}-e^{t}+c\) \(y \cdot e^{g(x)}=g(x) e^{g(x)}-e^{g(x)}+c\) \(e^{g(x)}(y-g(x)+1)=c\) Taking \(\log\) on both side- \(\mathrm{g}(\mathrm{x})+\log (1+\mathrm{y}-\mathrm{g}(\mathrm{x}))=\mathrm{c}\)
AP EAMCET-18.09.2020
Differential Equation
87300
If \(y=y(x)\) is the solution of the differential equation \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) with \(y(0)=1\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{4}{3}\)
Explanation:
(A) : We have differential equation- \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0\) \(\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}=-\cos x \Rightarrow \frac{d y}{y+1}=-\frac{\cos x}{2+\sin x} d x\) Integrating on both side- Given, \(\int \frac{d y}{y+1}=\int-\frac{\cos x}{2+\sin x} d x\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \(\log (y+1)=-\log (2+\sin x)+\log c\) \((y+1)(2+\sin x)=c\) \(y(0)=1\) \((1+1)(2+0)=c\) \(c=4\) Then equation become- \((y+1)(2+\sin x)=4\) Now, \(\mathrm{y}(\pi / 2)\) we get- \((y+1)(2+\sin \pi / 2)=4 \Rightarrow(y+1)(2+1)=4\) \((y+1)=\frac{4}{3} \Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\)
