87176
The general solution of the differential equation \(e^{\frac{d y}{d x}}=x\) is
1 \(y=x(\log x-1)+c\)
2 \(x=y(\log x-1)+c\)
3 \(x=y(\log x+1)+c\)
4 \(y=x(\log x+1)+c\)
Explanation:
(A) :We have differential equation- \(\mathrm{e}^{\frac{\mathrm{dy}}{\mathrm{dx}}}=\mathrm{X}\) Taking log on both sides, we get- \(\frac{d y}{d x} \log e=\log x \Rightarrow \quad \frac{d y}{d x}=\log x\) \(d y=\log x \cdot d x\) Integrating on both sides, we get- \(\int \mathrm{dy}=\int \log \mathrm{x} \cdot 1 \mathrm{dx}\) Apply integration by parts, \(y=\log x \int 1 \cdot d x-\int\left[(\log x)^{\prime}\left\{\int 1 d x\right\}\right] d x\) \(y=x \log x-\int\left(\frac{1}{x} \cdot x\right) d x \Rightarrow y=x \log x-x+c\) \(y=x(\log x-1)+c\)
MHT CET-2019
Differential Equation
87177
If \(r\) is the radius of spherical balloon at time \(t\) and the surface area of balloon changes at a constant rate \(k\), then
(B) : We know that, surface area of balloon with radius \(\mathrm{r}\) is \(4 \pi \mathrm{r}^{2}\). According to question, \(\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)=\mathrm{k} \Rightarrow 4 \pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k}\) \(8 \pi \mathrm{r} \mathrm{dr}=\mathrm{kdt}\) Now integrating both side - \(8 \pi \int \mathrm{rdr}=\mathrm{k} \int \mathrm{dt} \Rightarrow \quad 8 \pi \frac{\mathrm{r}^{2}}{2}=\mathrm{kt}+\mathrm{c}\) \(4 \pi \mathrm{r}^{2}=\mathrm{kt}+\mathrm{c}\)
MHT CET-2019
Differential Equation
87178
The solution of the differential equation \(\frac{d x}{d y}+\frac{1+x}{1-y}=0\) is
(B) : Given differential equation, \(\frac{d x}{d y}+\frac{1+x}{1-y}=0 \Rightarrow \quad \frac{d x}{d y}=-\frac{1+x}{1-y}\) \(\frac{d x}{1+x}+\frac{d y}{1-y}=0\) Integrating both sides we get- \(\log (1+x)-\log (1-y)=\log c\) \(\frac{1+x}{1-y}=c\)
MHT CET-2019
Differential Equation
87179
The differential equation of all lines having slope \(\mathrm{m}\), passing through origin is
1 \(\frac{d y}{d x}=m\)
2 \(\frac{d^{2} y}{d x^{2}}=0\)
3 \(x \frac{d y}{d x}=y\)
4 \(y=\frac{d y}{d x}+c\)
Explanation:
(A) : Given, differential equation of all lines having slope, \(\mathrm{m}\) \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) Passing through \((0,0)\) origin \(0=m \times 0+c\) \(\mathrm{c}=0\) Then, \(\mathrm{y}=\mathrm{mx}+0\) \(y=m x\) Differentiating on both sides w.r.t \(x\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\)
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Differential Equation
87176
The general solution of the differential equation \(e^{\frac{d y}{d x}}=x\) is
1 \(y=x(\log x-1)+c\)
2 \(x=y(\log x-1)+c\)
3 \(x=y(\log x+1)+c\)
4 \(y=x(\log x+1)+c\)
Explanation:
(A) :We have differential equation- \(\mathrm{e}^{\frac{\mathrm{dy}}{\mathrm{dx}}}=\mathrm{X}\) Taking log on both sides, we get- \(\frac{d y}{d x} \log e=\log x \Rightarrow \quad \frac{d y}{d x}=\log x\) \(d y=\log x \cdot d x\) Integrating on both sides, we get- \(\int \mathrm{dy}=\int \log \mathrm{x} \cdot 1 \mathrm{dx}\) Apply integration by parts, \(y=\log x \int 1 \cdot d x-\int\left[(\log x)^{\prime}\left\{\int 1 d x\right\}\right] d x\) \(y=x \log x-\int\left(\frac{1}{x} \cdot x\right) d x \Rightarrow y=x \log x-x+c\) \(y=x(\log x-1)+c\)
MHT CET-2019
Differential Equation
87177
If \(r\) is the radius of spherical balloon at time \(t\) and the surface area of balloon changes at a constant rate \(k\), then
(B) : We know that, surface area of balloon with radius \(\mathrm{r}\) is \(4 \pi \mathrm{r}^{2}\). According to question, \(\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)=\mathrm{k} \Rightarrow 4 \pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k}\) \(8 \pi \mathrm{r} \mathrm{dr}=\mathrm{kdt}\) Now integrating both side - \(8 \pi \int \mathrm{rdr}=\mathrm{k} \int \mathrm{dt} \Rightarrow \quad 8 \pi \frac{\mathrm{r}^{2}}{2}=\mathrm{kt}+\mathrm{c}\) \(4 \pi \mathrm{r}^{2}=\mathrm{kt}+\mathrm{c}\)
MHT CET-2019
Differential Equation
87178
The solution of the differential equation \(\frac{d x}{d y}+\frac{1+x}{1-y}=0\) is
(B) : Given differential equation, \(\frac{d x}{d y}+\frac{1+x}{1-y}=0 \Rightarrow \quad \frac{d x}{d y}=-\frac{1+x}{1-y}\) \(\frac{d x}{1+x}+\frac{d y}{1-y}=0\) Integrating both sides we get- \(\log (1+x)-\log (1-y)=\log c\) \(\frac{1+x}{1-y}=c\)
MHT CET-2019
Differential Equation
87179
The differential equation of all lines having slope \(\mathrm{m}\), passing through origin is
1 \(\frac{d y}{d x}=m\)
2 \(\frac{d^{2} y}{d x^{2}}=0\)
3 \(x \frac{d y}{d x}=y\)
4 \(y=\frac{d y}{d x}+c\)
Explanation:
(A) : Given, differential equation of all lines having slope, \(\mathrm{m}\) \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) Passing through \((0,0)\) origin \(0=m \times 0+c\) \(\mathrm{c}=0\) Then, \(\mathrm{y}=\mathrm{mx}+0\) \(y=m x\) Differentiating on both sides w.r.t \(x\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\)
87176
The general solution of the differential equation \(e^{\frac{d y}{d x}}=x\) is
1 \(y=x(\log x-1)+c\)
2 \(x=y(\log x-1)+c\)
3 \(x=y(\log x+1)+c\)
4 \(y=x(\log x+1)+c\)
Explanation:
(A) :We have differential equation- \(\mathrm{e}^{\frac{\mathrm{dy}}{\mathrm{dx}}}=\mathrm{X}\) Taking log on both sides, we get- \(\frac{d y}{d x} \log e=\log x \Rightarrow \quad \frac{d y}{d x}=\log x\) \(d y=\log x \cdot d x\) Integrating on both sides, we get- \(\int \mathrm{dy}=\int \log \mathrm{x} \cdot 1 \mathrm{dx}\) Apply integration by parts, \(y=\log x \int 1 \cdot d x-\int\left[(\log x)^{\prime}\left\{\int 1 d x\right\}\right] d x\) \(y=x \log x-\int\left(\frac{1}{x} \cdot x\right) d x \Rightarrow y=x \log x-x+c\) \(y=x(\log x-1)+c\)
MHT CET-2019
Differential Equation
87177
If \(r\) is the radius of spherical balloon at time \(t\) and the surface area of balloon changes at a constant rate \(k\), then
(B) : We know that, surface area of balloon with radius \(\mathrm{r}\) is \(4 \pi \mathrm{r}^{2}\). According to question, \(\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)=\mathrm{k} \Rightarrow 4 \pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k}\) \(8 \pi \mathrm{r} \mathrm{dr}=\mathrm{kdt}\) Now integrating both side - \(8 \pi \int \mathrm{rdr}=\mathrm{k} \int \mathrm{dt} \Rightarrow \quad 8 \pi \frac{\mathrm{r}^{2}}{2}=\mathrm{kt}+\mathrm{c}\) \(4 \pi \mathrm{r}^{2}=\mathrm{kt}+\mathrm{c}\)
MHT CET-2019
Differential Equation
87178
The solution of the differential equation \(\frac{d x}{d y}+\frac{1+x}{1-y}=0\) is
(B) : Given differential equation, \(\frac{d x}{d y}+\frac{1+x}{1-y}=0 \Rightarrow \quad \frac{d x}{d y}=-\frac{1+x}{1-y}\) \(\frac{d x}{1+x}+\frac{d y}{1-y}=0\) Integrating both sides we get- \(\log (1+x)-\log (1-y)=\log c\) \(\frac{1+x}{1-y}=c\)
MHT CET-2019
Differential Equation
87179
The differential equation of all lines having slope \(\mathrm{m}\), passing through origin is
1 \(\frac{d y}{d x}=m\)
2 \(\frac{d^{2} y}{d x^{2}}=0\)
3 \(x \frac{d y}{d x}=y\)
4 \(y=\frac{d y}{d x}+c\)
Explanation:
(A) : Given, differential equation of all lines having slope, \(\mathrm{m}\) \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) Passing through \((0,0)\) origin \(0=m \times 0+c\) \(\mathrm{c}=0\) Then, \(\mathrm{y}=\mathrm{mx}+0\) \(y=m x\) Differentiating on both sides w.r.t \(x\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\)
87176
The general solution of the differential equation \(e^{\frac{d y}{d x}}=x\) is
1 \(y=x(\log x-1)+c\)
2 \(x=y(\log x-1)+c\)
3 \(x=y(\log x+1)+c\)
4 \(y=x(\log x+1)+c\)
Explanation:
(A) :We have differential equation- \(\mathrm{e}^{\frac{\mathrm{dy}}{\mathrm{dx}}}=\mathrm{X}\) Taking log on both sides, we get- \(\frac{d y}{d x} \log e=\log x \Rightarrow \quad \frac{d y}{d x}=\log x\) \(d y=\log x \cdot d x\) Integrating on both sides, we get- \(\int \mathrm{dy}=\int \log \mathrm{x} \cdot 1 \mathrm{dx}\) Apply integration by parts, \(y=\log x \int 1 \cdot d x-\int\left[(\log x)^{\prime}\left\{\int 1 d x\right\}\right] d x\) \(y=x \log x-\int\left(\frac{1}{x} \cdot x\right) d x \Rightarrow y=x \log x-x+c\) \(y=x(\log x-1)+c\)
MHT CET-2019
Differential Equation
87177
If \(r\) is the radius of spherical balloon at time \(t\) and the surface area of balloon changes at a constant rate \(k\), then
(B) : We know that, surface area of balloon with radius \(\mathrm{r}\) is \(4 \pi \mathrm{r}^{2}\). According to question, \(\frac{\mathrm{d}}{\mathrm{dt}}\left(4 \pi \mathrm{r}^{2}\right)=\mathrm{k} \Rightarrow 4 \pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k}\) \(8 \pi \mathrm{r} \mathrm{dr}=\mathrm{kdt}\) Now integrating both side - \(8 \pi \int \mathrm{rdr}=\mathrm{k} \int \mathrm{dt} \Rightarrow \quad 8 \pi \frac{\mathrm{r}^{2}}{2}=\mathrm{kt}+\mathrm{c}\) \(4 \pi \mathrm{r}^{2}=\mathrm{kt}+\mathrm{c}\)
MHT CET-2019
Differential Equation
87178
The solution of the differential equation \(\frac{d x}{d y}+\frac{1+x}{1-y}=0\) is
(B) : Given differential equation, \(\frac{d x}{d y}+\frac{1+x}{1-y}=0 \Rightarrow \quad \frac{d x}{d y}=-\frac{1+x}{1-y}\) \(\frac{d x}{1+x}+\frac{d y}{1-y}=0\) Integrating both sides we get- \(\log (1+x)-\log (1-y)=\log c\) \(\frac{1+x}{1-y}=c\)
MHT CET-2019
Differential Equation
87179
The differential equation of all lines having slope \(\mathrm{m}\), passing through origin is
1 \(\frac{d y}{d x}=m\)
2 \(\frac{d^{2} y}{d x^{2}}=0\)
3 \(x \frac{d y}{d x}=y\)
4 \(y=\frac{d y}{d x}+c\)
Explanation:
(A) : Given, differential equation of all lines having slope, \(\mathrm{m}\) \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) Passing through \((0,0)\) origin \(0=m \times 0+c\) \(\mathrm{c}=0\) Then, \(\mathrm{y}=\mathrm{mx}+0\) \(y=m x\) Differentiating on both sides w.r.t \(x\), we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}\)