<p>pweease make an exception for just one calc question! i have the best intentions :) </p>
<p>i'm just stuck! it's probably easy, too..</p>
<p>Write the particular solution to the given differential equation with the initial condition f(1)=1.</p>
<p>I’m prettyyy sure you’re missing a part of that question, like, the “differential equation”?</p>
<p>i would forget that…</p>
<p>sorry i’m an idiot!</p>
<p>the differential equation is 1/2yy’-x=0 and let y=f(x) be a solution and there’s three parts to the question. the one i posted was part c.</p>
<p>i also need help with part b. haha it says for the particular solution with the initial condition f(2)=-1, write the equation of the tangent line to the graph of f at x=2. (can someone help me with this or do you need the graph? haha sorry i’m terrrrible at calc obviously )</p>
<p>Try using the Laplace Transform. :D</p>
<p>^The Laplace Transform (nor any other type of Fourier transform) is not necessary.</p>
<p>This is actually a very simple ordinary differential equation. Use the equation. Move things around a bit on both sides of the equation (you can treat y’ as dy/dx like a fraction). </p>
<p>PM me if you still need help.</p>
<p>Solution:
Given 1/2yy’-x=0 y(x)=f(x) f(1)=1
(1/2)y (dy/dx) – x =0
(1/2)y dy =x dx
Integrate the right and left side:
(1/4)y2 =(1/2) x2 + C where C is a constant
y(x)=(2 x2 +2C)0.5
y(x)=(2 x2 +C1)0.5
y(1)=(2 +C1)0.5=1 - > C1=-1
So y(x)= (2 x2 -1)0.5</p>
<p>Test the solution:
y(x)=(2 x2 +C1)0.5
y’(x)=(1/2)(2 x2 +C1)-0…5</p>
<p>(1/2) (2 x2 +C1)0.5 (1/2)(2 x2 +C1)-0…5 -x = 0
All done.
Best regards</p>