87216
If \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\), then \(f(x y)\) is equal to
1 \(k \cdot e^{\frac{x^{2}}{2}}\)
2 k.e \(\mathrm{y}^{\mathrm{y}^{2 / 2}}\)
3 \(k \cdot e^{x^{2}}\)
4 \(k . e^{\frac{x y}{2}}\)
Explanation:
(A) : We have \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\) The equation can be written as, i. e \(\quad \frac{d}{d x}(x y)=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y)}{f(x y)} d(x y)=x d x\) Integrating on both sides we get- \(\int \frac{f^{\prime}(x y)}{f(x y)} d(x y)=\int x d x \Rightarrow \log [f(x y)]=\frac{x^{2}}{2}+C\) \(f(x y)=e^{\frac{x^{2}}{2}+C}=e^{\frac{x^{2}}{2}} \cdot e^{C}\) \(f(x y)=k \cdot e^{\frac{x^{2}}{2}} \quad\left(\therefore e^{c}=k\right)\)
VITEEE-2014]**#
Differential Equation
87217
The solution of the differential equation \(\frac{d y}{d x}=(4 x+y+1)^{2}\), is
(C): The equation of the family of circles of radius \(r\) is- \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{i}\) Where \(\mathrm{a} \& \mathrm{~b}\) are arbitrary constants. Since equation (i) contains two arbitrary constants, we differentiate it two times \& the differential equation will be second order. Differentiating (i) w.r.t.x, we get \(2(x-a)+2(y-b) \frac{d y}{d x}=0\) \(\Rightarrow(\mathrm{x}-\mathrm{a})+(\mathrm{y}-\mathrm{b}) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) Differentiating (ii) w.r.t.x, we get \(1+(y-b) \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\) \(\Rightarrow(y-b)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\) On putting the value of \((y-b)\) in equation (ii), we get \(x-a=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d^{2}}}\) Substituting the values of \((x-a) \&(y-b)\) in (i), we get \(\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=r^{2}\) \(\Rightarrow\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\mathrm{r}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx})^{2}}\right.\)
87216
If \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\), then \(f(x y)\) is equal to
1 \(k \cdot e^{\frac{x^{2}}{2}}\)
2 k.e \(\mathrm{y}^{\mathrm{y}^{2 / 2}}\)
3 \(k \cdot e^{x^{2}}\)
4 \(k . e^{\frac{x y}{2}}\)
Explanation:
(A) : We have \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\) The equation can be written as, i. e \(\quad \frac{d}{d x}(x y)=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y)}{f(x y)} d(x y)=x d x\) Integrating on both sides we get- \(\int \frac{f^{\prime}(x y)}{f(x y)} d(x y)=\int x d x \Rightarrow \log [f(x y)]=\frac{x^{2}}{2}+C\) \(f(x y)=e^{\frac{x^{2}}{2}+C}=e^{\frac{x^{2}}{2}} \cdot e^{C}\) \(f(x y)=k \cdot e^{\frac{x^{2}}{2}} \quad\left(\therefore e^{c}=k\right)\)
VITEEE-2014]**#
Differential Equation
87217
The solution of the differential equation \(\frac{d y}{d x}=(4 x+y+1)^{2}\), is
(C): The equation of the family of circles of radius \(r\) is- \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{i}\) Where \(\mathrm{a} \& \mathrm{~b}\) are arbitrary constants. Since equation (i) contains two arbitrary constants, we differentiate it two times \& the differential equation will be second order. Differentiating (i) w.r.t.x, we get \(2(x-a)+2(y-b) \frac{d y}{d x}=0\) \(\Rightarrow(\mathrm{x}-\mathrm{a})+(\mathrm{y}-\mathrm{b}) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) Differentiating (ii) w.r.t.x, we get \(1+(y-b) \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\) \(\Rightarrow(y-b)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\) On putting the value of \((y-b)\) in equation (ii), we get \(x-a=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d^{2}}}\) Substituting the values of \((x-a) \&(y-b)\) in (i), we get \(\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=r^{2}\) \(\Rightarrow\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\mathrm{r}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx})^{2}}\right.\)
87216
If \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\), then \(f(x y)\) is equal to
1 \(k \cdot e^{\frac{x^{2}}{2}}\)
2 k.e \(\mathrm{y}^{\mathrm{y}^{2 / 2}}\)
3 \(k \cdot e^{x^{2}}\)
4 \(k . e^{\frac{x y}{2}}\)
Explanation:
(A) : We have \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\) The equation can be written as, i. e \(\quad \frac{d}{d x}(x y)=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y)}{f(x y)} d(x y)=x d x\) Integrating on both sides we get- \(\int \frac{f^{\prime}(x y)}{f(x y)} d(x y)=\int x d x \Rightarrow \log [f(x y)]=\frac{x^{2}}{2}+C\) \(f(x y)=e^{\frac{x^{2}}{2}+C}=e^{\frac{x^{2}}{2}} \cdot e^{C}\) \(f(x y)=k \cdot e^{\frac{x^{2}}{2}} \quad\left(\therefore e^{c}=k\right)\)
VITEEE-2014]**#
Differential Equation
87217
The solution of the differential equation \(\frac{d y}{d x}=(4 x+y+1)^{2}\), is
(C): The equation of the family of circles of radius \(r\) is- \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{i}\) Where \(\mathrm{a} \& \mathrm{~b}\) are arbitrary constants. Since equation (i) contains two arbitrary constants, we differentiate it two times \& the differential equation will be second order. Differentiating (i) w.r.t.x, we get \(2(x-a)+2(y-b) \frac{d y}{d x}=0\) \(\Rightarrow(\mathrm{x}-\mathrm{a})+(\mathrm{y}-\mathrm{b}) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) Differentiating (ii) w.r.t.x, we get \(1+(y-b) \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\) \(\Rightarrow(y-b)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\) On putting the value of \((y-b)\) in equation (ii), we get \(x-a=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d^{2}}}\) Substituting the values of \((x-a) \&(y-b)\) in (i), we get \(\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=r^{2}\) \(\Rightarrow\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\mathrm{r}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx})^{2}}\right.\)
87216
If \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\), then \(f(x y)\) is equal to
1 \(k \cdot e^{\frac{x^{2}}{2}}\)
2 k.e \(\mathrm{y}^{\mathrm{y}^{2 / 2}}\)
3 \(k \cdot e^{x^{2}}\)
4 \(k . e^{\frac{x y}{2}}\)
Explanation:
(A) : We have \(x \cdot \frac{d y}{d x}+y=x \cdot \frac{f(x y)}{f^{\prime}(x y)}\) The equation can be written as, i. e \(\quad \frac{d}{d x}(x y)=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y)}{f(x y)} d(x y)=x d x\) Integrating on both sides we get- \(\int \frac{f^{\prime}(x y)}{f(x y)} d(x y)=\int x d x \Rightarrow \log [f(x y)]=\frac{x^{2}}{2}+C\) \(f(x y)=e^{\frac{x^{2}}{2}+C}=e^{\frac{x^{2}}{2}} \cdot e^{C}\) \(f(x y)=k \cdot e^{\frac{x^{2}}{2}} \quad\left(\therefore e^{c}=k\right)\)
VITEEE-2014]**#
Differential Equation
87217
The solution of the differential equation \(\frac{d y}{d x}=(4 x+y+1)^{2}\), is
(C): The equation of the family of circles of radius \(r\) is- \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{i}\) Where \(\mathrm{a} \& \mathrm{~b}\) are arbitrary constants. Since equation (i) contains two arbitrary constants, we differentiate it two times \& the differential equation will be second order. Differentiating (i) w.r.t.x, we get \(2(x-a)+2(y-b) \frac{d y}{d x}=0\) \(\Rightarrow(\mathrm{x}-\mathrm{a})+(\mathrm{y}-\mathrm{b}) \frac{\mathrm{dy}}{\mathrm{dx}}=0\) Differentiating (ii) w.r.t.x, we get \(1+(y-b) \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\) \(\Rightarrow(y-b)=-\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\) On putting the value of \((y-b)\) in equation (ii), we get \(x-a=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right] \frac{d y}{d x}}{\frac{d^{2} y}{d^{2}}}\) Substituting the values of \((x-a) \&(y-b)\) in (i), we get \(\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}\left(\frac{d y}{d x}\right)^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}+\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{2}}{\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}=r^{2}\) \(\Rightarrow\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\mathrm{r}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx})^{2}}\right.\)
