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Question

(D) Given the differential equation $\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}$.This is a homogeneous differential equation. Let $y = vx$,then

Vedclass · Accepted Answer

$y = x \ln x + x$

Vedclass · Answer

(D) Given the differential equation $\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation,we get $v + x \frac{dv}{dx} = 1 + v$.Subtracting $v$ from both sides,we have $x \frac{dv}{dx} = 1$.Separating the variables,we get $dv = \frac{dx}{x}$.Integrating both sides,we get $v = \ln|x| + C$.Since $y = vx$,we have $\frac{y}{x} = \ln|x| + C$,which implies $y = x \ln|x| + Cx$.Given the condition $y(1) = 1$,we substitute $x = 1$ and $y = 1$:$1 = 1 \cdot \ln(1) + C(1) \Rightarrow 1 = 0 + C \Rightarrow C = 1$.Thus,the solution is $y = x \ln x + x$.

The solution of the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition $y(1) = 1$ is:

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