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Could you help me to explain how to find the solution of this equation $$y ′ (t)=−y(t)-\frac1{2}*(1+e^{-2t})+1$$ Given $y(0)=0$ Thank all This is my answer $$y ′ (t)=−y(t)-\frac1{2}e^{-2t}+\frac1{2}$$ $$e^{2t}y ′ (t)=e^{2t}(−y(t)-\frac1{2}e^{-2t}+\frac1{2})$$ where $$(e^{2t}y(t))′=e^{2t}y(t)′+2(e^{2t}y(t))=e^{2t}(y(t)-\frac1{2}e^{-2t}+\frac1{2})$$

user3051460
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5 Answers5

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Reordering, $$y'(t)+y(t)=\frac{1}{2}e^{-2t}+\frac{1}{2}$$

This DE is of the form $$y'(t)+P(t)y(t)=Q(t)$$ with $P(t)=1$ and $Q(t)=\frac{1}{2}e^{-2t}+\frac{1}{2}$.

Then, use the integrating factor $$M(t)=e^{\int_{t_0}^tP(x)\,dx}=e^{\int_{t_0}^t1\,dx}=e^{t-t_0}=e^{t-0}=e^t$$

So, in our DE, $$\begin{array}{rcl} M(t)y'(t)+M(t)y(t)&=&M(t)Q(t)\\ e^ty'(t)+e^ty(t)&=&e^t\bigg(\frac{1}{2}e^{-2t}+\frac{1}{2}\bigg)\\ \bigg(e^ty(t)\bigg)'&=&\frac{1}{2}\big(e^{t}+e^{-t}\big)\\ e^ty(t)&=&\frac{1}{2}\int_{t_0}^t\big(e^{t}+e^{-t}\big)\,dt\\ e^ty(t)&=&\frac{1}{2} \big(e^t-e^{-t}\big)\bigg|_{0}^{t}\\ y(t)&=&\frac{1}{2e^t} \big(e^{t-0}-e^{-(t-0)}\big)\\ y(t)&=&\frac{1}{2e^t} \big(e^{t}-e^{-t}\big)\\ \end{array}$$

Why use the integrating factor? Note that $M'(t)=P(t)M(t)$, so multiplying the integrating factor in the DE gives: $$M(t)y'(t)+M(t)y(t)=M(t)Q(t)\Rightarrow \bigg(M(t)y(t)\bigg)'=M(t)Q(t)$$ So the solution is: $$y(t)=\frac{1}{M(t)}\int_{t_0}^tM(x)Q(x)\,dx=\frac{1}{e^{\int_{t_0}^tP(x)\,dx}}\int_{t_0}^te^{\int_{t_0}^xP(s)\,ds}Q(x)\,dx$$

cjferes
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  • I think you missing C const in the solution. But your explain is perfect and the question is -1/2 – user3051460 Oct 06 '14 at 15:43
  • Note that I use the initial condition, using $t_0=0$ when integrating. If you prefer, I can edit the answer so you take the initial condition after the calculation of the general solution... – cjferes Oct 06 '14 at 15:46
  • Also, the $-1/2$ is multiplying a $1$, and you sum a $1$ afterwards.. do the math! – cjferes Oct 06 '14 at 15:47
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Multiply the equation by $e^t$, then $$ e^t(y'+y)=\frac{e^t}{2}-\frac{e^{-t}}{2} $$ or $$ \big(e^t y\big)'=\frac{1}{2}\big(e^t+e^{-t}\big)' $$ or $$ e^t y=\frac{1}{2}\big(e^t+e^{-t}\big)+c, $$ for some $c$ constant, or $$ y=\frac{1}{2}\big(1+e^{-2t}\big)+c\,e^{-t}. $$

Yiorgos S. Smyrlis
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one solution of the equation $y'(t)+y(t)=0$ is $y=C_1e^{-t}$

Dr. Sonnhard Graubner
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Hints; Solve $$y'+y=\frac12 \left (1-e^{-2t} \right ) \stackrel{\cdot e^{\int 1 \mathrm{d}t}}{\iff} e^t y'+e^t y= e^t \frac12 \left (1-e^{-2t} \right )$$

Do you recognise the LHS as something familiar?

UserX
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Work it out assuming y = u*v, and y' = u'v + uv'.

Rearrange to y' + y = 1/2 - e-2t/2

Solve homogeneous equation

u' + u = 0

So u = C * e-t.

Now substitute for y in full equation

u'v + uv' + uv = 1/2 - e-2t/2

v*(u' + u) + u*v' = 1/2 - e-2t/2

0 + u*v' = 1/2 - e-2t/2

v' = et/2 - e-t/2

Integrate

v = et/2 + e-t/2

y = Auv + Bu

= A*(1/2 + e-2t/2) + B*e-t

Checking y' + y = -1/2(1 + e-2t/2 + 1 , find A = 1, and B is arbitrary.

y = (1/2 + e-2t/2) + B*e-t

Paul Magnussen
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