(A) : We have differential equation- \(\frac{d y}{d x}+y=2 e^{2 x}\) Which is of the form, \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=1, \mathrm{Q}=2 \mathrm{e}^{\mathrm{e}}\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 1 \mathrm{dx}}=\mathrm{e}^{\mathrm{x}}\) For general solution we get- \(y(\text { I.F })=\int I \cdot F \times Q d x+C\) \(y^{x}=\int e^{x} \cdot 2 e^{2 x} d x+C \Rightarrow y e^{x}=2 \int e^{3 x} d x\) \(y e^{x}=2 \cdot \frac{e^{3 x}}{3}+C\) \(y=\frac{2}{3} e^{2 x}+C e^{-x}=C e^{-x}+\frac{2}{3} e^{2 x}\)
JCECE-2007
Differential Equation
87243
The solution of differential equation \((x+y)(d x\) - dy) \(=d x+d y\) is:
(D) : We have differential equation- \((x+y)(d x-d y)=d x+d y\) \(d x-d y=\frac{d x+d y}{x+y} \Rightarrow \quad d(x-y)=\frac{d(x+y)}{x+y}\) Integrating on both side, we get- \(\int \mathrm{d}(\mathrm{x}-\mathrm{y})=\int \frac{\mathrm{d}(\mathrm{x}+\mathrm{y})}{(\mathrm{x}+\mathrm{y})}+\mathrm{C}\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y})+\log C\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y}) \cdot \mathrm{C} \Rightarrow \mathrm{e}^{\mathrm{x}-\mathrm{y}}=C \cdot(\mathrm{x}+\mathrm{y})\) \(\mathrm{x}+\mathrm{y}=\frac{1}{\mathrm{C}} \mathrm{e}^{\mathrm{x}-\mathrm{y}}\) \(\mathrm{x}+\mathrm{y}=\mathrm{ke}^{\mathrm{x}-\mathrm{y}} \quad\left(\because \mathrm{k}=\frac{1}{C}\right)\)
JCECE-2005
Differential Equation
87244
The differential equation \(x \frac{d y}{d x}-y=x^{3}\) has the general solution
1 \(y-x^{3}=2 C x\)
2 \(2 y-x^{3}=2 C x\)
3 \(2 y+x^{2}=2 \mathrm{Cx}\)
4 \(y+x^{2}=2 C x\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}-y=x^{3} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{2}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) Where, \(P=-1 / x \text { and } Q=x^{2}\) \(I . F=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) Now general solution, we get- \(y \cdot(\text { I.F })=\int I . F \times Q d x+C\) \(y \frac{1}{x}=\int_{3} \frac{1}{x} x^{2} d x+C \Rightarrow \quad \frac{y}{x}=\frac{x^{2}}{2}+C\) \(2 y=x^{3}+2 C x\) \(2 y-x^{3}=2 C x\)
(A) : We have differential equation- \(\frac{d y}{d x}+y=2 e^{2 x}\) Which is of the form, \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=1, \mathrm{Q}=2 \mathrm{e}^{\mathrm{e}}\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 1 \mathrm{dx}}=\mathrm{e}^{\mathrm{x}}\) For general solution we get- \(y(\text { I.F })=\int I \cdot F \times Q d x+C\) \(y^{x}=\int e^{x} \cdot 2 e^{2 x} d x+C \Rightarrow y e^{x}=2 \int e^{3 x} d x\) \(y e^{x}=2 \cdot \frac{e^{3 x}}{3}+C\) \(y=\frac{2}{3} e^{2 x}+C e^{-x}=C e^{-x}+\frac{2}{3} e^{2 x}\)
JCECE-2007
Differential Equation
87243
The solution of differential equation \((x+y)(d x\) - dy) \(=d x+d y\) is:
(D) : We have differential equation- \((x+y)(d x-d y)=d x+d y\) \(d x-d y=\frac{d x+d y}{x+y} \Rightarrow \quad d(x-y)=\frac{d(x+y)}{x+y}\) Integrating on both side, we get- \(\int \mathrm{d}(\mathrm{x}-\mathrm{y})=\int \frac{\mathrm{d}(\mathrm{x}+\mathrm{y})}{(\mathrm{x}+\mathrm{y})}+\mathrm{C}\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y})+\log C\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y}) \cdot \mathrm{C} \Rightarrow \mathrm{e}^{\mathrm{x}-\mathrm{y}}=C \cdot(\mathrm{x}+\mathrm{y})\) \(\mathrm{x}+\mathrm{y}=\frac{1}{\mathrm{C}} \mathrm{e}^{\mathrm{x}-\mathrm{y}}\) \(\mathrm{x}+\mathrm{y}=\mathrm{ke}^{\mathrm{x}-\mathrm{y}} \quad\left(\because \mathrm{k}=\frac{1}{C}\right)\)
JCECE-2005
Differential Equation
87244
The differential equation \(x \frac{d y}{d x}-y=x^{3}\) has the general solution
1 \(y-x^{3}=2 C x\)
2 \(2 y-x^{3}=2 C x\)
3 \(2 y+x^{2}=2 \mathrm{Cx}\)
4 \(y+x^{2}=2 C x\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}-y=x^{3} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{2}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) Where, \(P=-1 / x \text { and } Q=x^{2}\) \(I . F=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) Now general solution, we get- \(y \cdot(\text { I.F })=\int I . F \times Q d x+C\) \(y \frac{1}{x}=\int_{3} \frac{1}{x} x^{2} d x+C \Rightarrow \quad \frac{y}{x}=\frac{x^{2}}{2}+C\) \(2 y=x^{3}+2 C x\) \(2 y-x^{3}=2 C x\)
(A) : We have differential equation- \(\frac{d y}{d x}+y=2 e^{2 x}\) Which is of the form, \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=1, \mathrm{Q}=2 \mathrm{e}^{\mathrm{e}}\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 1 \mathrm{dx}}=\mathrm{e}^{\mathrm{x}}\) For general solution we get- \(y(\text { I.F })=\int I \cdot F \times Q d x+C\) \(y^{x}=\int e^{x} \cdot 2 e^{2 x} d x+C \Rightarrow y e^{x}=2 \int e^{3 x} d x\) \(y e^{x}=2 \cdot \frac{e^{3 x}}{3}+C\) \(y=\frac{2}{3} e^{2 x}+C e^{-x}=C e^{-x}+\frac{2}{3} e^{2 x}\)
JCECE-2007
Differential Equation
87243
The solution of differential equation \((x+y)(d x\) - dy) \(=d x+d y\) is:
(D) : We have differential equation- \((x+y)(d x-d y)=d x+d y\) \(d x-d y=\frac{d x+d y}{x+y} \Rightarrow \quad d(x-y)=\frac{d(x+y)}{x+y}\) Integrating on both side, we get- \(\int \mathrm{d}(\mathrm{x}-\mathrm{y})=\int \frac{\mathrm{d}(\mathrm{x}+\mathrm{y})}{(\mathrm{x}+\mathrm{y})}+\mathrm{C}\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y})+\log C\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y}) \cdot \mathrm{C} \Rightarrow \mathrm{e}^{\mathrm{x}-\mathrm{y}}=C \cdot(\mathrm{x}+\mathrm{y})\) \(\mathrm{x}+\mathrm{y}=\frac{1}{\mathrm{C}} \mathrm{e}^{\mathrm{x}-\mathrm{y}}\) \(\mathrm{x}+\mathrm{y}=\mathrm{ke}^{\mathrm{x}-\mathrm{y}} \quad\left(\because \mathrm{k}=\frac{1}{C}\right)\)
JCECE-2005
Differential Equation
87244
The differential equation \(x \frac{d y}{d x}-y=x^{3}\) has the general solution
1 \(y-x^{3}=2 C x\)
2 \(2 y-x^{3}=2 C x\)
3 \(2 y+x^{2}=2 \mathrm{Cx}\)
4 \(y+x^{2}=2 C x\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}-y=x^{3} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{2}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) Where, \(P=-1 / x \text { and } Q=x^{2}\) \(I . F=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) Now general solution, we get- \(y \cdot(\text { I.F })=\int I . F \times Q d x+C\) \(y \frac{1}{x}=\int_{3} \frac{1}{x} x^{2} d x+C \Rightarrow \quad \frac{y}{x}=\frac{x^{2}}{2}+C\) \(2 y=x^{3}+2 C x\) \(2 y-x^{3}=2 C x\)
(A) : We have differential equation- \(\frac{d y}{d x}+y=2 e^{2 x}\) Which is of the form, \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=1, \mathrm{Q}=2 \mathrm{e}^{\mathrm{e}}\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 1 \mathrm{dx}}=\mathrm{e}^{\mathrm{x}}\) For general solution we get- \(y(\text { I.F })=\int I \cdot F \times Q d x+C\) \(y^{x}=\int e^{x} \cdot 2 e^{2 x} d x+C \Rightarrow y e^{x}=2 \int e^{3 x} d x\) \(y e^{x}=2 \cdot \frac{e^{3 x}}{3}+C\) \(y=\frac{2}{3} e^{2 x}+C e^{-x}=C e^{-x}+\frac{2}{3} e^{2 x}\)
JCECE-2007
Differential Equation
87243
The solution of differential equation \((x+y)(d x\) - dy) \(=d x+d y\) is:
(D) : We have differential equation- \((x+y)(d x-d y)=d x+d y\) \(d x-d y=\frac{d x+d y}{x+y} \Rightarrow \quad d(x-y)=\frac{d(x+y)}{x+y}\) Integrating on both side, we get- \(\int \mathrm{d}(\mathrm{x}-\mathrm{y})=\int \frac{\mathrm{d}(\mathrm{x}+\mathrm{y})}{(\mathrm{x}+\mathrm{y})}+\mathrm{C}\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y})+\log C\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y}) \cdot \mathrm{C} \Rightarrow \mathrm{e}^{\mathrm{x}-\mathrm{y}}=C \cdot(\mathrm{x}+\mathrm{y})\) \(\mathrm{x}+\mathrm{y}=\frac{1}{\mathrm{C}} \mathrm{e}^{\mathrm{x}-\mathrm{y}}\) \(\mathrm{x}+\mathrm{y}=\mathrm{ke}^{\mathrm{x}-\mathrm{y}} \quad\left(\because \mathrm{k}=\frac{1}{C}\right)\)
JCECE-2005
Differential Equation
87244
The differential equation \(x \frac{d y}{d x}-y=x^{3}\) has the general solution
1 \(y-x^{3}=2 C x\)
2 \(2 y-x^{3}=2 C x\)
3 \(2 y+x^{2}=2 \mathrm{Cx}\)
4 \(y+x^{2}=2 C x\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}-y=x^{3} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{2}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) Where, \(P=-1 / x \text { and } Q=x^{2}\) \(I . F=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) Now general solution, we get- \(y \cdot(\text { I.F })=\int I . F \times Q d x+C\) \(y \frac{1}{x}=\int_{3} \frac{1}{x} x^{2} d x+C \Rightarrow \quad \frac{y}{x}=\frac{x^{2}}{2}+C\) \(2 y=x^{3}+2 C x\) \(2 y-x^{3}=2 C x\)
(A) : We have differential equation- \(\frac{d y}{d x}+y=2 e^{2 x}\) Which is of the form, \(\frac{d y}{d x}+P y=Q\) \(\mathrm{P}=1, \mathrm{Q}=2 \mathrm{e}^{\mathrm{e}}\) I.F \(=\mathrm{e}^{\int P \mathrm{dx}}=\mathrm{e}^{\int 1 \mathrm{dx}}=\mathrm{e}^{\mathrm{x}}\) For general solution we get- \(y(\text { I.F })=\int I \cdot F \times Q d x+C\) \(y^{x}=\int e^{x} \cdot 2 e^{2 x} d x+C \Rightarrow y e^{x}=2 \int e^{3 x} d x\) \(y e^{x}=2 \cdot \frac{e^{3 x}}{3}+C\) \(y=\frac{2}{3} e^{2 x}+C e^{-x}=C e^{-x}+\frac{2}{3} e^{2 x}\)
JCECE-2007
Differential Equation
87243
The solution of differential equation \((x+y)(d x\) - dy) \(=d x+d y\) is:
(D) : We have differential equation- \((x+y)(d x-d y)=d x+d y\) \(d x-d y=\frac{d x+d y}{x+y} \Rightarrow \quad d(x-y)=\frac{d(x+y)}{x+y}\) Integrating on both side, we get- \(\int \mathrm{d}(\mathrm{x}-\mathrm{y})=\int \frac{\mathrm{d}(\mathrm{x}+\mathrm{y})}{(\mathrm{x}+\mathrm{y})}+\mathrm{C}\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y})+\log C\) \(\mathrm{x}-\mathrm{y}=\log (\mathrm{x}+\mathrm{y}) \cdot \mathrm{C} \Rightarrow \mathrm{e}^{\mathrm{x}-\mathrm{y}}=C \cdot(\mathrm{x}+\mathrm{y})\) \(\mathrm{x}+\mathrm{y}=\frac{1}{\mathrm{C}} \mathrm{e}^{\mathrm{x}-\mathrm{y}}\) \(\mathrm{x}+\mathrm{y}=\mathrm{ke}^{\mathrm{x}-\mathrm{y}} \quad\left(\because \mathrm{k}=\frac{1}{C}\right)\)
JCECE-2005
Differential Equation
87244
The differential equation \(x \frac{d y}{d x}-y=x^{3}\) has the general solution
1 \(y-x^{3}=2 C x\)
2 \(2 y-x^{3}=2 C x\)
3 \(2 y+x^{2}=2 \mathrm{Cx}\)
4 \(y+x^{2}=2 C x\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}-y=x^{3} \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{2}\) Which is the form of \(\frac{d y}{d x}+P y=Q\) Where, \(P=-1 / x \text { and } Q=x^{2}\) \(I . F=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\) Now general solution, we get- \(y \cdot(\text { I.F })=\int I . F \times Q d x+C\) \(y \frac{1}{x}=\int_{3} \frac{1}{x} x^{2} d x+C \Rightarrow \quad \frac{y}{x}=\frac{x^{2}}{2}+C\) \(2 y=x^{3}+2 C x\) \(2 y-x^{3}=2 C x\)
