Question

Q.4
Solve the differential equation (x² + y²)dy/dx=xy​

Answer 1

Answer:

$\frac{ {x}^{2} }{2 {y}^{2} } - log |y| + 2log |x| = C\\ \:$

Step-by-step explanation:

To Solve the differential equation,by inspection it is clear it can't be solved by separating variables.

So, it can be a homogeneous differential equation,let us try to convert the entire equation in y/x form

$( {x}^{2} + {y}^{2} ) \frac{dy}{dx} = xy \\ \\ {x}^{2} (1 + \frac{ {y}^{2} }{ {x}^{2} } ) \frac{dy}{dx} = xy \\ \\ (1 + \frac{ {y}^{2} }{ {x}^{2} } ) \frac{dy}{dx} = \frac{y}{x} \\ \\ \frac{dy}{dx} = \frac{ \frac{y}{x} }{1 + \frac{ {y}^{2} }{ {x}^{2} } } =f\big(\frac{y}{x}\big)$

put

$\frac{y}{x} = v \\ \\ y = xv \\ \\ \frac{dy}{dx} = v + x \frac{dv}{dx} \\ \\ v + x \frac{dv}{dx} = \frac{ v }{1 + {v}^{2} } \\ \\ x \frac{dv}{dx} = \frac{ v }{1 + {v}^{2} } - v \\ \\ x \frac{dv}{dx} = \frac{ v - v - {v}^{3} }{1 + {v}^{2} }\\ \\ x \frac{dv}{dx} =\frac{ - {v}^{3} }{1 + {v}^{2} }\\ \\ separate \: the \: variables \\ \\ - \frac{1 + {v}^{2} }{ {v}^{ 3} } dv = \frac{1}{x} dx \\ \\ integrate \: the \: above \: equation\\ \\ - \int\frac{1 + {v}^{2} }{ {v}^{ 3} } dv = \int\frac{1}{x} dx \\ \\ - \int\frac{1}{ {v}^{ 3} } dv - \int \frac{1}{v}dv = \int\frac{1}{x} dx \\ \\ apply \: power \: rule \: of \: integration \\ \\ \frac{1}{2 {v}^{2} } - log \: |v| = - log |x| + C \\ \\ \frac{1}{2 {v}^{2} } - log \: |v| + log |x| = C$

Now undo substitution

$\frac{1}{2 {( \frac{y}{x} })^{2} } - log \: | \frac{y}{x} | + log |x| = C\\ \\ \frac{ {x}^{2} }{2 {y}^{2} } - log |y| + log |x| + log |x| = C\\ \\ \frac{ {x}^{2} }{2 {y}^{2} } - log |y| + 2log |x| = C \\ \\or\\\\ \frac{ {x}^{2} }{2 {y}^{2} } + log \Big|\frac{x^2}{y}\Big| = C$

is the required solution.

Hope it helps you.

Answer 2

$-\frac{x^2}{2y^2}+ln y=C$

Step-by-step explanation:

$(x^2+y^2)\frac{dy}{dx}=xy$

$\frac{dy}{dx}=\frac{xy}{x^2+y^2}$

Let y=vx

Differentiate w.r.t x

$\frac{dy}{dx}=x\frac{dv}{dx}+v$

Substitute the values

$v+x\frac{dv}{dx}=\frac{vx^2}{v^2x^2+x^2}$

$x\frac{dv}{dx}=\frac{vx^2}{x^2(1+v^2)}-v$

$x\frac{dv}{dx}=\frac{v}{1+v^2}-v=\frac{v-v-v^3}{1+v^2}$

$\frac{dv}{dx}=\frac{-v^3}{x(1+v^2)}$

$\frac{1+v^2}{v^3}dv=-\frac{1}{x}dx$

$\int(\frac{1}{v^3}+\frac{1}{v})dv=-\int\frac{dx}{x}$

$\int(v^{-3}+\frac{1}{v})dv=-\int\frac{dx}{x}$

$\frac{v^{-2}}{-2}+ln v=-ln x+C$

Using the formula

$\int x^ndx=\frac{x^{n+1}}{n+1}$

$\int\frac{dx}{x}=ln x$

$-\frac{1}{2v^2}+ln v+ln x=C$

$-\frac{1}{2v^2}+ln vx=C$

Using the formula

$lnx+ln y=ln xy$

Substitute the values

$-\frac{x^2}{2y^2}+ln y=C$

#Learn more:

https://brainly.in/question/5815907:Answered by Mathdude