Let y=y(x) be the solution curve of the diffe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the differential equation: $(y^{2}-x) \frac{dy}{dx}=1$.Rearranging the equation,we get: $\frac{dx}{dy} = y^{2}-x$,which implies $\frac{dx}{d

Vedclass · Accepted Answer

$2-e$

Vedclass · Answer

(C) Given the differential equation: $(y^{2}-x) \frac{dy}{dx}=1$.Rearranging the equation,we get: $\frac{dx}{dy} = y^{2}-x$,which implies $\frac{dx}{dy} + x = y^{2}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y)=1$ and $Q(y)=y^{2}$.The integrating factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int 1 dy} = e^{y}$.The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$.$x e^{y} = \int y^{2} e^{y} dy + C$.Using integration by parts for $\int y^{2} e^{y} dy$: $\int y^{2} e^{y} dy = y^{2} e^{y} - \int 2y e^{y} dy = y^{2} e^{y} - 2(y e^{y} - e^{y}) = (y^{2}-2y+2)e^{y}$.So,$x e^{y} = (y^{2}-2y+2)e^{y} + C$.Given the condition $y(0)=1$,we substitute $x=0$ and $y=1$:$0 \cdot e^{1} = (1^{2}-2(1)+2)e^{1} + C \Rightarrow 0 = (1)e + C \Rightarrow C = -e$.The equation of the curve is $x e^{y} = (y^{2}-2y+2)e^{y} - e$.To find the intersection with the $x$-axis,set $y=0$:$x e^{0} = (0^{2}-2(0)+2)e^{0} - e \Rightarrow x(1) = 2(1) - e \Rightarrow x = 2-e$.

Let $y=y(x)$ be the solution curve of the differential equation $(y^{2}-x) \frac{dy}{dx}=1$ satisfying $y(0)=1$. This curve intersects the $x$-axis at a point whose abscissa is

Similar Questions

The general solution of the differential equation,$y' + y\phi'(x) - \phi(x)\phi'(x) = 0$,where $\phi(x)$ is a known function,is: (where $c$ is an arbitrary constant)

The integrating factor of the differential equation $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$ is

The general solution of the differential equation,$\sin 2x \left( \frac{dy}{dx} - \sqrt{\tan x} \right) - y = 0$ is

If $\log y$ is an integrating factor of $\frac{dx}{dy} + P(y)x = Q(y)$,then $P(y) =$

The solution of $\frac{dy}{dx} + y = e^x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module