Thread: Specific Solution for a Differential Equation

1. Specific Solution for a Differential Equation

Find the solution of the diff eq that satisfies the given initial condition.

y'tan x = a + y, y(pi/4) = a, 0 < x < pi/2

y' = (a+y)/tanx

dy/ (a+y) = cot x dx

Integrate...

ln (a + y ) = ln (sinx ) + c

I think to make this work, c will be 2a- root2/2, so my answer is
looks like

y = sinx - a + 2a - root2 /2

Is this correct?

2. Originally Posted by veronicak5678 Find the solution of the diff eq that satisfies the given initial condition.

y'tan x = a + y, y(pi/4) = a, 0 < x < pi/2

y' = (a+y)/tanx

dy/ (a+y) = cot x dx

Integrate...

ln (a + y ) = ln (sinx ) + c

I think to make this work, c will be 2a- root2/2, so my answer is
looks like

y = sinx - a + 2a - root2 /2

Is this correct?

When solving for y, you should end up with $\displaystyle a+y=e^{\ln(\sin x)+C}\implies y=C\sin x-a$

When you apply the initial condition, you should get $\displaystyle C=2\sqrt{2}a$

Thus, $\displaystyle \color{red}\boxed{y=2\sqrt{2}a\sin x-a}$

Does this make sense?

--Chris

3. No... could you show me how to solve for y?

4. Originally Posted by veronicak5678 No... could you show me how to solve for y?
Your work is good up to here:

$\displaystyle \ln(a+y)=\ln(\sin x)+C$

To get rid of the logarithm on the left side of the equation, apply the exponential function:

$\displaystyle \ln(a+y)=\ln(\sin x)+C\implies e^{\ln(a+y)}=e^{\ln(\sin x)+C}\implies a+y=e^{\ln(\sin x)+C}$

Thus,

$\displaystyle a+y=e^{\ln(\sin x)+C}\implies a+y=e^{\ln(\sin x)}e^C\implies a+y=Ce^{\ln(\sin x)}$; $\displaystyle e^C=C$

Finally,

$\displaystyle a+y=Ce^{\ln(\sin x)}\implies a+y=C\sin x\implies y=C\sin x-a$

Now plug in the initial condition to find C.

Does this clarify things?

--Chris

5. Got it. I would never have known to do that! Thanks a lot.

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