Question

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

y=

Answer 1

The solution for the initial value problem is:

`y(x) = -5e^{(x^(2))/(2) - 2x} + 10`

Explanation:

We have the following initial value problem:

`(dy)/(dx) = (x-2)(y-10)`

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:

`(dy)/(dx) = (x-2)(y-10)`

`(dy)/(y-10) = (x-2)dx`

To find y in function of x, we integrate both sides.

`(dy)/(y-10) = (x-2)dx`

`\int {(1)/(y-10)} \, dy = \int {(x-2)} \, dx`

Solving each integral separately

`\int {(1)/(y-10)} \, dy`

This one we solve by substitution

`u = y-10, du = dy`

`\int {(1)/(y-10)} \, dy = \int {(1)/(u)} \, du = ln(u) = ln((y-10))`

`\int {(x-2)} \, dx = (x^(2))/(2) - 2x + K`

Now we have that:

`\int {(1)/(y-10)} \, dy = \int {(x-2)} \, dx`

`\ln{(y-10) = (x^(2))/(2) - 2x + K}`

To solve for y, we apply the exponential to both sides, since the exponential and ln are inverse operations:

`e^{ln((y-10)) = e^{(x^(2))/(2) - 2x + K}`

`y - 10 = Ke^{(x^(2))/(2) - 2x}`

`y(x) = Ke^{(x^(2))/(2) - 2x} + 10`

`y(0) = 5` means that when
`x = 0, y(x) = 5`. So:

`5 = Ke^{(0^(2))/(2) - 2*0} + 10`

`Ke^(0) = -5`

`K = -5`

So, the solution for the initial value problem is:

`y(x) = -5e^{(x^(2))/(2) - 2x} + 10`