Question

yy'=x^3+y^2/x,y(2)=6 find the initial value solution of the differential equations.​

Answer 1

Step-by-step explanation:

We have,

$\tt{y \cdot y^{ \prime} = \dfrac{ {x}^{3} + {y}^{2} }{x} }$

$\tt{ \implies\dfrac{dy}{dx} = \dfrac{ {x}^{3} + {y}^{2} }{xy} }$

$\tt{ \implies\dfrac{dy}{dx} = \dfrac{ {x}^{2} }{y} + \dfrac{ y }{x} }$

$\tt{ \implies\dfrac{dy}{dx} -\dfrac{ y }{x} = \dfrac{ {x}^{2} }{y} }$

$\tt{ \implies \: y\dfrac{dy}{dx} -\dfrac{ {y}^{2} }{x} = {x}^{2} }$

$\bf{Put \: \: \: y^{2} = v }$

$\bf{ \implies \: 2y \dfrac{dy}{dx} = \dfrac{dv}{dx} }$

$\bf{ \implies \: y \dfrac{dy}{dx} = \dfrac{1}{2} \cdot \dfrac{dv}{dx} }$

So,

$\tt{ \implies \: \dfrac{1}{2} \cdot\dfrac{dv}{dx} -\dfrac{v }{x} = {x}^{2} }$

$\tt{ \implies \: \dfrac{dv}{dx} -\dfrac{2v }{x} = 2 {x}^{2} }$

This is linear form, so,

$\displaystyle\bf{ I.F. = {e}^{ \displaystyle \bf{ - \int \dfrac{2}{x} \: dx}} }$

$\displaystyle \implies\bf{ I.F. = {e}^{ \displaystyle \bf{ - 2\int \dfrac{dx}{x} }} }$

$\displaystyle \implies\bf{ I.F. = {e}^{ \displaystyle \bf{ - 2 \ln | x | }} }$

$\displaystyle \implies\bf{ I.F. = {e}^{ \displaystyle \bf{ \ln | x |^{ - 2} }} }$

$\displaystyle \implies\bf{ I.F. = \dfrac{1}{ x ^{ 2}} }$

So, our solution is

$\displaystyle\tt{v \cdot \dfrac{1}{ {x}^{2} } = \int2 {x}^{2} \cdot \dfrac{1}{ {x}^{2} } \: dx }$

$\displaystyle\tt{ \implies \dfrac{v}{ {x}^{2} } = \int2 \: dx }$

$\displaystyle\tt{ \implies \dfrac{v}{ {x}^{2} } = 2\int \: dx }$

$\tt{ \implies \dfrac{v}{ {x}^{2} } = 2x + c }$

Put the value of v,

$\tt{ \implies \dfrac{ {y}^{2} }{ {x}^{2} } = 2x + c }$

Since y(2) = 6,

So,

$\tt{ \implies \dfrac{(6)^{2} }{ (2)^{2} } = 2(2) + c }$

$\tt{ \implies \dfrac{36}{ 4} = 4 + c }$

$\tt{ \implies 9 - 4 = c }$

$\tt{ \implies c = 5 }$

So, required solution,

$\tt{ \implies \dfrac{ {y}^{2} }{ {x}^{2} } = 2x + 5 }$

$\implies \boxed{\tt{ {y}^{2} = 2x^{3} + 5 {x}^{2} }}$