Solve the following differential equation (dy/dx)=x+5y, y(0)=3?
dy/dx = x + 5y cannot be separated. I assume you got: −5y dy = x dx But this is incorrect. When moving 5y to left side you would get: dy/dx = x + 5y −5y + dy/dx = x −5y dx + dy = x dx So this will not work. Instead we write in the form: dy/dx + p(x) y = q(x) then find an integrating factor. dy/dx = x + 5y dy/dx − 5y = x p(x) = −5, q(x) = x Integrating factor = e^(∫ p(x) dx) = e^(∫−5 dx) = e^(−5x) Multiply both sides of differential by e^(−5x) e^(−5x) dy/dx − 5e^(−5x) y = x e^(−5x) Left side is = to derivative of e^(−5x) y with respect to x d/dx (e^(−5x) y) = x e^(−5x) d (e^(−5x) y) = x e^(−5x) dx Integrate both sides (use integration by parts on right side) ∫ d (e^(−5x) y) = ∫ x e^(−5x) dx e^(−5x) y = −1/25 e^(−5x) (5x + 1) + C y = −1/25 (5x + 1) + C e^(5x) y(0) = 3 3 = −1/25 (0 + 1) + C C = 76/25 y = 1/25 (76 e^(5x) − 5x − 1) —————————————————————————————— Check: x + 5y = x + 5 * 1/25 (76 e^(5x) − 5x − 1) x + 5y = x + 1/5 (76 e^(5x) − 5x − 1) x + 5y = 1/5 (5x + 76 e^(5x) − 5x − 1) x + 5y = 1/5 (76 e^(5x) − 1) dy/dx = 1/25 (5 * 76 e^(5x) − 5) dy/dx = 1/5 (76 e^(5x) − 1) dy/dx = x + 5y OK
Solve the differential equation dy/dx = (3e^(2x) + 3e^(4x)) /e^x + e^(-x)?
Dangi, your final answer is wrong. it should be e^(3x). input your formula 3e^(3x)into http://www.mathHandbook.com, click the integrate button for solution.reference math handbookhttp://www.mathHandbook.com
How can I solve the differential equation dy/dx = e^ (3x-2y)?
How can I solve the differential equation dy/dx = e^ (3x-2y)?There is already an answer, but it’s hard to read. So here’s another answer.Separation of variables is often worth a try. Multiplying both sides by [math]e^{2y}dx[/math] gives [math]e^{2y}dy=e^{3x}dx[/math] (*) and you can now solve by quadratures (integrating both sides) giving [math]\frac12e^{2y}=\frac13(e^{3x}+c)[/math].You can easily write [math]y[/math] as a function of [math]x[/math]: [math]2y-\ln(2)=\ln(e^{3x}+c)-\ln(3)[/math]. In the special case that [math]c=0[/math], [math]y = \frac32\ln\left(\frac23\right)x[/math].*Multiplying by [math]dx[/math] seems wrong as [math]dx[/math] is not a number, but it works. You can justify it by substitution, but that complicates the issue.
Solve the differential equation: dy/dx=1-x+y-xy?
dy/dx=1-x+y-xy dy/dx = (1- x) + y(1 - x) dy/dx = (1 - x)(1 + y) ... dy ∫---------- = (1 - x)dx ...1 + y ln(1 + y) = x - 1/2x^2 + C 1 + y = e^(x - 1/2x^2) e^C y = Ce^(x - 1/2x^2) - 1 answer//
What is the solution to this differential equation: sec(x)dy/dx=e^(sin(x)+y)?
Step 1: Clarify domain[math]sec(x)[/math] is undefined for [math]x=\frac{\pi}{2}+n\pi[/math] where [math]n\in\mathbb{N}[/math]Save this information until laterStep 2: Solve, assuming we are in an allowed region[math]sec(x)\frac{dy}{dx}=e^{sin(x)+y}[/math][math]sec(x)\frac{dy}{dx}=e^{sin(x)}e^y[/math][math]e^{-y}\frac{dy}{dx}=cos(x)e^{sin(x)}[/math][math]\int_\alpha^ye^{-t}dt = \int_\alpha^xcos(t)e^{sin(t)}dt[/math][math]-e^{-y}+e^{-\alpha}=e^{sin(x)}-e^{sin(\alpha)}[/math][math]e^{-y}=C-e^{sin(x)}[/math] where [math]C=e^{-\alpha}+e^{sin(\alpha)}[/math][math]y=-ln|C-e^{sin(x)}|[/math]Step 3: Details, details, details….Notice that the differential equations “can be written”[math]\frac{dy}{dx}=cos(x)e^{sin(x)+y}[/math]Since the right-hand side is continuous, there exists a unique solution to this differential equations.HOWEVER, this is not the same equation! We lose the discontinuities at [math]x=\frac{pi}{2}+n\pi[/math]So first, let’s just confirm that our solution still works:[math]\frac{d}{dx}(-ln|C-e^{sin(x)}|)=\frac{cos(x)e^{sin(x)}}{C-e^{sin(x)}}[/math][math]sec(x)\frac{dy}{dx}=\frac{sec(x)cos(x)e^{sin(x)}}{C-e^{sin(x)}}[/math][math]sec(x)\frac{dy}{dx}=\frac{e^{sin(x)}{e^{-y}}[/math][math]sec(x)\frac{dy}{dx}=e^{sin(x)+y}[/math]Great, so we still have “a solution!”In fact, as long as the solution is defined, we don’t care about the formula for C!Let’s say we have the above differential equation as part of an IVP with [math]y(x_0)=y_0[/math]. Then we can solve for C:[math]y_0=-ln|C-e^{sin(x_0)}|[/math][math]C=e^{sin(x_0)}+\frac{1}{y_0}[/math]So it looks like any value of C is allowed. However, this will restrict the domain of the function since if [math]sin(x)=ln(C)[/math], the solution blows up.Moreover, at one of the discontinuities of [math]sec(x)[/math], the left-hand side of the differential equation is no longer defined, so discontinuities in the solution should probably be allowed (I can’t find any good references on this, so I may be wrong). Thus, the most general solution to the differential is equation is [math]y=-ln|C-e^{sin(x)}|[/math] where the constant of integration can change when crossing [math]x=\frac{\pi}{2}+n\pi[/math] (unsure about that last part)
What is the solution of the differential equation (dy/dx) ^2-xdy/dx+y=0?
(dy/dx)^2 -x(dy/dx)+y=0representing dy/dx by D we have:D^2 - xD+y=0Or, y=xD-D^2Or, dy/dx=xdD/dx +D-2DdD/dxOr, 0=xdD/dx-2DdD/dxdD/dx=0;x-2D=0So, x=2dy/dxInteg xdx= Integ 2 dyOr, x^2 /2 =2yOr, y=x^2/4+ k
What is the answer of differential equation d^2y/dx^2 + 2 dy/dx + y = 0 when x=0, y=1, dy/dx = 2?
d^2y/dx^2 + 2 dy/dx + y = 0 when x=0, y=1, dy/dx = 2. Substitute the given values into the equation…( tilt your phone for this). Still, ( refer to the picture because I'm in a hurry at the moment and thus can't afford to type much)EDIT: now that I've learnt integral calculus, let's jump in!A sketch of the question.Warning: terrible handwriting ahead321Part 1Also,Clearing things up…Final answer.
Find the general solution of the differential equation dy/dx+2y=e^x?
dy/dx+2y=e^x P(x) = 2, Q(x) = e^(x) IF = e^[2∫dx] = e^(2x) ye^(2x) = ∫ e^(2x) * e^(x)dx ye^(2x) =∫ e^(3x)dx ye^(2x) = 1/3e^(3x) + C y = 1/3e^(x) + Ce^(-2x) answer//
What is the general solution of the differential equation (2x+2y+3) dy/dx= (x+y+1)?
[math]\dfrac {dy}{x+y+1}=\dfrac {dx}{2x+2y+3}=\dfrac {dy+dx}{x+y+1+2x+2y+3}=\dfrac {d(y+x)}{3(x+y)+4}[/math]Also [math]\dfrac {dy}{x+y+1}=\dfrac {dx}{2x+2y+3}=\dfrac {dx-2 dy}{2x+2y+3–2(x+y+1)}=\dfrac {d(x-2 y)}{1}[/math][math]\therefore \dfrac {d(y+x)}{3(x+y)+4}=\dfrac {d(x-2 y)}{1}[/math]Integrating,[math]\boxed{ \dfrac 1 3 \ln|3(x+y)+4|=x-2 y+C}[/math]Be short ;Stand tall.