Question

Solve dy/dx=xy with initial condition y(0)=1.

Answer 1

it is a separable ordinary differential equation. first step is to separate the y terms and x terms.

\[\frac{ dy }{ y } = x dx\]

next step is integrating both sides:

\[\int\limits_{}^{} \frac{ dy }{ y } = \int\limits_{}^{}x dx\]

\[\ln|y| + C = \frac{ x^2 }{ 2 } + C\]

Both C's are arbitrary so you can add them and put them both on one side.

\[\ln|y| = \frac{ x^2 }{ 2 } + C\]

next step is to solve for y, wanna do it?

Answer 2

y=e^(x^2/2), right?

Answer 3

yes, but actually:

\[y = e^{(\frac{ x^2 }{ 2 } + C)} = e^{\frac{ x^2 }{ 2 }}e^{C} \]

Since e^C can be any real positive number, we can say that e^C = C (an arbitrary constant)
, giving:
\[y = Ce^{\frac{ x^2 }{ 2 } }\]

this is the general solution for the form of differential equation that was solved "dy/dx = xy"

the value of C depends on the initial condition. which in this case is y(0) = 1

plugging that in the general solution:

\[y(0) = 1 = Ce^{0} = C\] C = 1 for this initial condition.

Therefore the final solution is:

\[y(x) = e^{\frac{ x^2 }{ 2 }}\]

Answer 4

So the answer is y=e^(x^2/2).

Answer 5

yes. sorry for delay lol

Answer 6

$$\frac{dy}{dx}=xy\\\frac1ydy=x\,dx\\\int\frac1ydy=\int x\,dx\\\log y=\frac12x^2+C\\y=Ce^{\frac12x^2}$$

Answer 7

$$y(0)=1\\Ce^0=1\\C=1$$so we know our particular solution is \(y=e^{\frac12 x^2}\)

Answer 8

Thank you so much.