Q:

Solve the initial value problem dydx=(x−2)(y−10),y(0)=5dydx=(x−2)(y−10),y(0)=5.y=

Accepted Solution

A:
Answer:The solution for the initial value problem is:[tex]y(x) = -5e^{\frac{x^{2}}{2} - 2x} + 10[/tex]Step-by-step explanation:We have the following initial value problem:[tex]\frac{dy}{dx} = (x-2)(y-10)[/tex]The first step is solving the differential equation. We can do this by the variable separation method. It means that every term with y in on one side of the equality, every term with x on the other side. So:[tex]\frac{dy}{dx} = (x-2)(y-10)[/tex][tex]\frac{dy}{y-10} = (x-2)dx[/tex]To find y in function of x, we integrate both sides.[tex]\frac{dy}{y-10} = (x-2)dx[/tex][tex]\int {\frac{1}{y-10}} \, dy = \int {(x-2)} \, dx[/tex]Solving each integral separately[tex]\int {\frac{1}{y-10}} \, dy[/tex]This one we solve by substitution[tex]u = y-10, du = dy[/tex][tex]\int {\frac{1}{y-10}} \, dy = \int {\frac{1}{u}} \, du = \ln{u} = \ln{(y-10)}[/tex][tex] \int {(x-2)} \, dx = \frac{x^{2}}{2} - 2x + K[/tex]Now we have that:[tex]\int {\frac{1}{y-10}} \, dy = \int {(x-2)} \, dx[/tex][tex]\ln{(y-10) = \frac{x^{2}}{2} - 2x + K}[/tex]To solve for y, we apply the exponential to both sides, since the exponential and ln are inverse operations:[tex]e^{\ln{(y-10)} = e^{\frac{x^{2}}{2} - 2x + K}[/tex][tex]y - 10 = Ke^{\frac{x^{2}}{2} - 2x}[/tex][tex]y(x) = Ke^{\frac{x^{2}}{2} - 2x} + 10[/tex][tex]y(0) = 5[/tex] means that when [tex]x = 0, y(x) = 5[/tex]. So:[tex]5 = Ke^{\frac{0^{2}}{2} - 2*0} + 10[/tex][tex]Ke^{0} = -5[/tex][tex]K = -5[/tex]So, the solution for the initial value problem is:[tex]y(x) = -5e^{\frac{x^{2}}{2} - 2x} + 10[/tex]