(A) : Given that, differential equation, \(x d y-y d x+x^{2} e^{x} d x=0\) \(x d y=y d x-x^{2} e^{x} d x\) \(\frac{d y}{d x}=\frac{y}{x}-\frac{x^{2} e^{x}}{x} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=-x e^{x}\) This is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-\frac{1}{\mathrm{x}}, \mathrm{Q}=-\mathrm{xe}^{\mathrm{x}} \Rightarrow \mathrm{I} \cdot F=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log \mathrm{x}}\) I. \(F=\frac{1}{\mathrm{X}}\) For general solution y (I.F) I.F \(\quad \mathrm{Qdx} \quad \mathrm{c} \Rightarrow \mathrm{y} \frac{1}{\mathrm{x}} \quad \mathrm{xe}^{\mathrm{x}} \frac{1}{\mathrm{x}} \mathrm{dx} \quad \mathrm{c}\) \(\underline{\mathrm{y}}=-\mathrm{e}^{\mathrm{x}}+\mathrm{c}\) \(\mathrm{x}\) \(\underline{y}+e^{x}=c\)
BCECE-2006
Differential Equation
87248
The differential equation of the family of curves \(y=a \cos (x+b)\) is :
1 \(\frac{d^{2} y}{d x^{2}}-y=0\)
2 \(\frac{d^{2} y}{d x^{2}}+y=0\)
3 \(\frac{d^{2} y}{d x^{2}}+2 y=0\)
4 none of these
Explanation:
(B) : We have differential equation \(y=a \cos (x+b)\) Differentiating w.r.t \(x\), we get- \(\frac{d y}{d x}=-a \sin (x+b)\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2} y}{d x^{2}}=-a \cos (x+b) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+\mathrm{y}=0\)
(A) : Given that, differential equation, \(x d y-y d x+x^{2} e^{x} d x=0\) \(x d y=y d x-x^{2} e^{x} d x\) \(\frac{d y}{d x}=\frac{y}{x}-\frac{x^{2} e^{x}}{x} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=-x e^{x}\) This is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-\frac{1}{\mathrm{x}}, \mathrm{Q}=-\mathrm{xe}^{\mathrm{x}} \Rightarrow \mathrm{I} \cdot F=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log \mathrm{x}}\) I. \(F=\frac{1}{\mathrm{X}}\) For general solution y (I.F) I.F \(\quad \mathrm{Qdx} \quad \mathrm{c} \Rightarrow \mathrm{y} \frac{1}{\mathrm{x}} \quad \mathrm{xe}^{\mathrm{x}} \frac{1}{\mathrm{x}} \mathrm{dx} \quad \mathrm{c}\) \(\underline{\mathrm{y}}=-\mathrm{e}^{\mathrm{x}}+\mathrm{c}\) \(\mathrm{x}\) \(\underline{y}+e^{x}=c\)
BCECE-2006
Differential Equation
87248
The differential equation of the family of curves \(y=a \cos (x+b)\) is :
1 \(\frac{d^{2} y}{d x^{2}}-y=0\)
2 \(\frac{d^{2} y}{d x^{2}}+y=0\)
3 \(\frac{d^{2} y}{d x^{2}}+2 y=0\)
4 none of these
Explanation:
(B) : We have differential equation \(y=a \cos (x+b)\) Differentiating w.r.t \(x\), we get- \(\frac{d y}{d x}=-a \sin (x+b)\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2} y}{d x^{2}}=-a \cos (x+b) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+\mathrm{y}=0\)
(A) : Given that, differential equation, \(x d y-y d x+x^{2} e^{x} d x=0\) \(x d y=y d x-x^{2} e^{x} d x\) \(\frac{d y}{d x}=\frac{y}{x}-\frac{x^{2} e^{x}}{x} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=-x e^{x}\) This is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-\frac{1}{\mathrm{x}}, \mathrm{Q}=-\mathrm{xe}^{\mathrm{x}} \Rightarrow \mathrm{I} \cdot F=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log \mathrm{x}}\) I. \(F=\frac{1}{\mathrm{X}}\) For general solution y (I.F) I.F \(\quad \mathrm{Qdx} \quad \mathrm{c} \Rightarrow \mathrm{y} \frac{1}{\mathrm{x}} \quad \mathrm{xe}^{\mathrm{x}} \frac{1}{\mathrm{x}} \mathrm{dx} \quad \mathrm{c}\) \(\underline{\mathrm{y}}=-\mathrm{e}^{\mathrm{x}}+\mathrm{c}\) \(\mathrm{x}\) \(\underline{y}+e^{x}=c\)
BCECE-2006
Differential Equation
87248
The differential equation of the family of curves \(y=a \cos (x+b)\) is :
1 \(\frac{d^{2} y}{d x^{2}}-y=0\)
2 \(\frac{d^{2} y}{d x^{2}}+y=0\)
3 \(\frac{d^{2} y}{d x^{2}}+2 y=0\)
4 none of these
Explanation:
(B) : We have differential equation \(y=a \cos (x+b)\) Differentiating w.r.t \(x\), we get- \(\frac{d y}{d x}=-a \sin (x+b)\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2} y}{d x^{2}}=-a \cos (x+b) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+\mathrm{y}=0\)
(A) : Given that, differential equation, \(x d y-y d x+x^{2} e^{x} d x=0\) \(x d y=y d x-x^{2} e^{x} d x\) \(\frac{d y}{d x}=\frac{y}{x}-\frac{x^{2} e^{x}}{x} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=-x e^{x}\) This is the form of \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=-\frac{1}{\mathrm{x}}, \mathrm{Q}=-\mathrm{xe}^{\mathrm{x}} \Rightarrow \mathrm{I} \cdot F=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log \mathrm{x}}\) I. \(F=\frac{1}{\mathrm{X}}\) For general solution y (I.F) I.F \(\quad \mathrm{Qdx} \quad \mathrm{c} \Rightarrow \mathrm{y} \frac{1}{\mathrm{x}} \quad \mathrm{xe}^{\mathrm{x}} \frac{1}{\mathrm{x}} \mathrm{dx} \quad \mathrm{c}\) \(\underline{\mathrm{y}}=-\mathrm{e}^{\mathrm{x}}+\mathrm{c}\) \(\mathrm{x}\) \(\underline{y}+e^{x}=c\)
BCECE-2006
Differential Equation
87248
The differential equation of the family of curves \(y=a \cos (x+b)\) is :
1 \(\frac{d^{2} y}{d x^{2}}-y=0\)
2 \(\frac{d^{2} y}{d x^{2}}+y=0\)
3 \(\frac{d^{2} y}{d x^{2}}+2 y=0\)
4 none of these
Explanation:
(B) : We have differential equation \(y=a \cos (x+b)\) Differentiating w.r.t \(x\), we get- \(\frac{d y}{d x}=-a \sin (x+b)\) Again differentiating w.r.t \(x\), we get- \(\frac{d^{2} y}{d x^{2}}=-a \cos (x+b) \Rightarrow \frac{d^{2} y}{d x^{2}}=-y\) \(\frac{d^{2} y}{d x^{2}}+\mathrm{y}=0\)
