The particular solution of the differential e | vedclass

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Question

(B) Given differential equation is $x dy + 2y dx = 0$. Dividing by $xy$,we get $\frac{dy}{y} + \frac{2 dx}{x} = 0$. Integrating both sides,we get $\in

Vedclass · Accepted Answer

$x^2 y = 4$

Vedclass · Answer

(B) Given differential equation is $x dy + 2y dx = 0$. Dividing by $xy$,we get $\frac{dy}{y} + \frac{2 dx}{x} = 0$. Integrating both sides,we get $\int \frac{dy}{y} + 2 \int \frac{dx}{x} = \int 0$. This gives $\ln|y| + 2 \ln|x| = C_1$. Using the property of logarithms,$\ln|y| + \ln|x^2| = C_1$,which implies $\ln|yx^2| = C_1$. Thus,$yx^2 = e^{C_1} = C$. Given $x = 2$ and $y = 1$,we substitute these values: $(1)(2)^2 = C$,so $C = 4$. Therefore,the particular solution is $x^2 y = 4$.

The particular solution of the differential equation $x dy + 2y dx = 0$,when $x = 2, y = 1$ is

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