Question

Y sin 2x dx =(1+y^+cos^x)dy

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y\cdot\,sin(2x)\,dx=\left(1+y^2+cos^2(x)\right)dy}$

$\tt{\implies\,sin(2x)\cdot\dfrac{dx}{dy}=\dfrac{1+y^2+cos^2(x)}{y}}$

$\tt{\implies\,sin(2x)\cdot\dfrac{dx}{dy}=\dfrac{1+y^2}{y}+\dfrac{cos^2(x)}{y}}$

$\tt{\implies\,sin(2x)\cdot\dfrac{dx}{dy}-\dfrac{cos^2(x)}{y}=\dfrac{1+y^2}{y}}$

$\bf{Put\,\,\,-cos^2(x)=v}$

$\bf{\implies-2\,cos(x)\cdot(-sin(x))\cdot\dfrac{dx}{dy}=\dfrac{dv}{dy}}$

$\bf{\implies2\,cos(x)\,sin(x)\cdot\dfrac{dx}{dy}=\dfrac{dv}{dy}}$

$\bf{\implies\,sin(2x)\cdot\dfrac{dx}{dy}=\dfrac{dv}{dy}}$

So,

$\tt{\implies\,\dfrac{dv}{dy}+\dfrac{v}{y}=\dfrac{1+y^2}{y}}$

This is linear form, so,

$\sf{I.F.=e^{\displaystyle\,\int\dfrac{dy}{y}}}$

$\sf{\implies\,I.F.=e^{\displaystyle\ln(y)}}$

$\sf{\implies\,I.F.=y}$

Now,

$\tt{\displaystyle\,v\cdot\,y=\int\dfrac{1+y^2}{y}\cdot\,y\,dy}$

$\tt{\displaystyle\implies\,v\cdot\,y=\int\left(1+y^2\right)dy}$

$\tt{\displaystyle\implies\,v\cdot\,y=y+\dfrac{y^3}{3}+c}$

Put the value of v

$\tt{\displaystyle\implies\,-cos^2(x)\cdot\,y=y+\dfrac{y^3}{3}+c}$

$\tt{\displaystyle\implies\,-3\,cos^2(x)\cdot\,y=3y+y^3+3c}$

$\tt{\displaystyle\implies\,3y+3\,cos^2(x)\cdot\,y+y^3+3c=0}$

$\tt{\displaystyle\implies\,3\,y\left(1+cos^2(x)\right)+y^3=-3c}$

Put -3c = C

$\tt{\displaystyle\implies\,3\,y\left(1+cos^2(x)\right)+y^3=C}$