If frac{dx}{dy} = frac{1+x-y^2}{y} and x(1) = | vedclass

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

Question

(A) Given the differential equation: $\frac{dx}{dy} = \frac{1+x-y^2}{y}$.Rearranging the terms,we get: $\frac{dx}{dy} - \frac{x}{y} = \frac{1-y^2}{y}$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given the differential equation: $\frac{dx}{dy} = \frac{1+x-y^2}{y}$.Rearranging the terms,we get: $\frac{dx}{dy} - \frac{x}{y} = \frac{1-y^2}{y}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = \frac{1-y^2}{y}$.The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.Substituting the values: $x \cdot \frac{1}{y} = \int \left(\frac{1-y^2}{y}\right) \cdot \frac{1}{y} dy + C$.$x \cdot \frac{1}{y} = \int \left(\frac{1}{y^2} - 1\right) dy + C$.$x \cdot \frac{1}{y} = -\frac{1}{y} - y + C$.Multiplying by $y$: $x = -1 - y^2 + Cy$.Given the condition $x(1) = 1$,we substitute $y=1$ and $x=1$: $1 = -1 - (1)^2 + C(1) \Rightarrow 1 = -2 + C \Rightarrow C = 3$.Thus,the particular solution is $x = -1 - y^2 + 3y$.To find $5x(2)$,we substitute $y=2$: $x(2) = -1 - (2)^2 + 3(2) = -1 - 4 + 6 = 1$.Therefore,$5x(2) = 5(1) = 5$.

If $\frac{dx}{dy} = \frac{1+x-y^2}{y}$ and $x(1) = 1$,then $5x(2)$ is equal to:

Similar Questions

The general solution of the differential equation $\left(\frac{1}{x^2}+x\right) \frac{d y}{d x}+3 y=1$ is

The solution of the equation $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$,where $y(0) = 1$,is

Let $y=y(x)$ be the solution of the differential equation $x^3 dy + (xy - 1) dx = 0, x > 0$,with $y(\frac{1}{2}) = 3 - e$. Then $y(1)$ is equal to

If for $x \geq 0$,$y=y(x)$ is the solution of the differential equation $(x+1) dy = ((x+1)^{2} + y - 3) dx$ with $y(2) = 0$,then $y(3)$ is equal to:

The integrating factor of the differential equation $x \frac{dy}{dx} + 2y = x^2$ is . . . . . . . $(x \neq 0)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module