Question #325141

x(dy)/(dx)=x^(2) + 5y

1
Expert's answer
2022-04-07T17:26:45-0400

xdydx=x2+5yx\frac{dy}{dx}=x^2+5y

y5yx=xy’-5\frac{y}{x}=x (1)

Let’s solve the following equation 

y5yx=0y’-5\frac{y}{x}=0

dyy=5dxx\frac {dy}{y}=5\frac{dx}{x}

y=C(x)x5y=C(x)x^{5}

To find solution of the equation (1) we should think C is a function of x

y=Cx5+5Cx4y'=C'x^{5}+5Cx^{4}

Substitution y and y’ into (1) gives

Cx5+5Cx45Cx5x=xC'x^{5}+5Cx^{4} -5\frac{Cx^{5}}{x}=x

C=x4C'=x^{-4}

C=13x3+C1C=\frac{1}{-3x^3}+C_1

y=(13x3+C1)x5=13x2+C1x5y=(\frac{1}{-3x^3}+C_1)x^5=-\frac13x^2+C_1x^5

Answer: y=(13x3+C1)x5=13x2+C1x5y=(\frac{1}{-3x^3}+C_1)x^5=-\frac13x^2+C_1x^5 .


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