Q:

dy/dx = (sin x)/y , y(0) = 2

Accepted Solution

A:
Answer:The solution for this differential equation is [tex]y=\sqrt{-2cos(x)+6}[/tex]Step-by-step explanation:This differential equation [tex]\frac{dy}{dx}=\frac{sin(x)}{y}[/tex] is a separable First-Order ordinary differential equation.We know this because a first-order differential equation is separable if and only if it can be written as[tex]\frac{dy}{dx}=f(x)g(y)[/tex] where f and g are known functions.And we have[tex]\frac{dy}{dx}=\frac{sin(x)}{y}\\ \frac{dy}{dx}=sin(x)\frac{1}{y}[/tex]To solve this differential equation we need to integrate both sides[tex]y\cdot dy=sin(x)\cdot dx\\ \int\limits {y\cdot dy}= \int\limits {sin(x)\cdot dx}[/tex][tex]\int\limits {y\cdot dy}=\frac{y^{2} }{2} + C[/tex][tex]\int\limits {sin(x) \cdot dx}=-cos(x) + C[/tex][tex]\frac{y^{2} }{2} + C=-cos(x) + C[/tex]We can make a new constant of integration [tex]C_{1}[/tex][tex]\frac{y^{2} }{2}=-cos(x) + C_{1}[/tex]We need to isolate y[tex]\frac{y^{2} }{2}=-cos(x) + C_{1}\\y^2=-2cos(x)+2*C_{1}\\\mathrm{For\:}y^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}y=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\y=\sqrt{-2cos(x)+c_{1} } \\y=-\sqrt{-2cos(x)+c_{1} }[/tex]We have the initial conditions y(0)=2 so we can find the value of the constant of integration for [tex]y=\sqrt{-2cos(x)+c_{1} } [/tex][tex]2=\sqrt{-2\cos \left(0\right)+c_1}\\2= \sqrt{-2+c_1} \\c_1=6[/tex]For [tex]y=-\sqrt{-2cos(x)+c_{1} } [/tex] there is not solution for [tex]c_{1}[/tex] in the domain of real numbers.The solution for this differential equation is [tex]y=\sqrt{-2cos(x)+6}[/tex]