The solution of the differential equation (x^ | vedclass

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Question

(A) Given differential equation: $(x^2+y^2) dx = 5xy dy$ $\Rightarrow \frac{dy}{dx} = \frac{x^2+y^2}{5xy}$ This is a homogeneous differential equation

Vedclass · Accepted Answer

$|x^2-4y^2|^5=x^2$

Vedclass · Answer

(A) Given differential equation: $(x^2+y^2) dx = 5xy dy$ $\Rightarrow \frac{dy}{dx} = \frac{x^2+y^2}{5xy}$ This is a homogeneous differential equation. Put $y = Vx$,then $\frac{dy}{dx} = V + x \frac{dV}{dx}$. Substituting these into the equation: $V + x \frac{dV}{dx} = \frac{x^2 + V^2x^2}{5x(Vx)} = \frac{1+V^2}{5V}$ $\Rightarrow x \frac{dV}{dx} = \frac{1+V^2}{5V} - V = \frac{1+V^2-5V^2}{5V} = \frac{1-4V^2}{5V}$ Separating variables: $\int \frac{5V}{1-4V^2} dV = \int \frac{dx}{x}$ Let $1-4V^2 = t$,then $-8V dV = dt$,so $V dV = -\frac{1}{8} dt$. $\Rightarrow 5 \int \frac{-1/8}{t} dt = \int \frac{dx}{x}$ $\Rightarrow -\frac{5}{8} \ln|t| = \ln|x| + C_1$ $\Rightarrow -5 \ln|1-4V^2| = 8 \ln|x| + C_2$ $\Rightarrow \ln|1-4V^2|^{-5} = \ln|x^8| + C_2$ $\Rightarrow |1-4V^2|^{-5} = K x^8$ $\Rightarrow |1-4(\frac{y}{x})^2|^{-5} = K x^8$ $\Rightarrow |\frac{x^2-4y^2}{x^2}|^{-5} = K x^8$ $\Rightarrow |x^2-4y^2|^{-5} \cdot (x^2)^5 = K x^8$ $\Rightarrow |x^2-4y^2|^{-5} = K x^{-2}$ $\Rightarrow |x^2-4y^2|^5 = C x^2$ Given $y(1)=0$: $|1^2 - 4(0)^2|^5 = C(1)^2 \Rightarrow C = 1$. Thus,the solution is $|x^2-4y^2|^5 = x^2$.

The solution of the differential equation $(x^2+y^2) dx - 5xy dy = 0$,$y(1)=0$,is :

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