Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the initial value problem $y' + 5y = 5t, y(0) = y_0$. Find the value of $y_0$ for which the solution touches, but does not cross, the $t$-axis.

This is linear first order DE, so I found integrating factor=$e^{(2/3)t}$ Solving ODE I got, $y=\frac{21}{8}-\frac{3}{4}t+Ce^{(-2/3)t}$ Using, $y(0)=y_0$, we get $C=y_0-\frac{21}{8}$. Now, if we take first derivative of the solution function $y$ to find the point where it has zero slope, we end up with two unknown in the equation, $y_0$ and $t$. $y'=\frac{-3}{4}+\frac{-2}{3}(y_0-\frac{21}{8})e^{-(2/3)t}=0.$

share|improve this question

3 Answers 3

First, your solution is incorrect. $$y'(t) + 5y(t) = 5t \implies \exp(5t) y'(t) +5y(t) \exp(5t) = 5t \exp(5t)\\ \implies \dfrac{d \left( y(t) \exp(5t)\right)}{dt} = 5t \exp(5t)$$ Hence, $$y(t) \exp(5t) = \left( t - \dfrac15 \right) \exp(5t) + c \implies y(t) = t - \dfrac15 + c \exp(-5t)$$ The slope is given by $$y'(t) = 5 (t-y(t)) = 5 \left(\dfrac15 - c \exp(-5t) \right) = 1 - 5c \exp(-5t)$$ For this to touch but not cross the $t$ axis, we need $y(t_0) = 0$ and $y'(t_0) = 0$. These give us $$t_0 - \dfrac15 + c \exp(-5t_0) = 0$$ and $$1 - 5c \exp(-5t_0) = 0$$ Now obtain $t_0$ and $c$ by solving these two equations. This gives us $t_0 = 0$ and $c = \dfrac15$. Hence, the curve is given by $$y(t) = t - \dfrac15 + \dfrac{\exp(-5t)}5.$$ Now $$y_0 = y(0) = 0$$

share|improve this answer
    
@Vikram: This is what you wanted. :) –  Babak S. Jul 1 '12 at 7:01
    
Actually I was afraid to say that I made a mistake while posting the question, the question that I have posted and my partial answer are both different, I also don't know from where that wrong DE got pasted here, sorry friends. :( –  Vikram Jul 1 '12 at 10:24

It is not necessary to solve the DE to answer this question.

Suppose that $y(t)$ is a solution to the DE that touches, but does not cross, the $t$-axis, and let $t_0$ denote the time at which this happens.

Since the solution $y$ touches the $t$-axis at the time $t_0$, we have $y(t_0) = 0$.

Since the solution $y$ does not cross the $t$-axis at this time, we must have $y'(t_0) = 0$. (If $y'(t_0)$ were positive, then there would be some $\delta > 0$ with the property that $y(t)$ is negative for all $t$ in $(t_0 - \delta, t_0)$ and positive for all $t$ in $(t_0, t_0 + \delta)$, and hence the solution $y$ crosses the $t$-axis instead of just touching touch it. The analysis is similar if $y'(t_0) < 0$.)

Since $y$ solves the DE, we have $$ y'(t_0) + 5 y(t_0) = 5 t_0. $$ Using the facts established earlier that $y'(t_0) = y(t_0) = 0$, we deduce from the above equation that $0 = 5 t_0$, and hence that $t_0 = 0$. It follows that $$ y(0) = y(t_0) = 0. $$ [To make the argument in "if $y'(t_0)$ were positive..." more rigorous, you need to observe that if $y$ solves the DE, then $y$ is differentiable, and hence continuous, and hence $y' = 5t - 5y$ is continuous. So the continuous function $y'$, being positive at $t_0$, must also be positive in a neighborhood of $t_0$. The claims about the sign of $y$ then follow from the mean value theorem.]

share|improve this answer

Your achieved solution is wrong! $$y'+5y=5t\rightarrow\mu(t)=\exp(\int5dt)=e^{5t}$$ $$\rightarrow d(e^{5t}y)=5te^{5t}\rightarrow y=t-\frac{1}{5}+Ce^{-5t}$$ if $y(0)=y_0$ $$y=25t-25+(y_0+\frac{1}{5})e^{-5t}$$ Now, take this latter one as you answer.

share|improve this answer
    
Nice correction! Thank goodness, for the OP's sake! +1 –  amWhy Mar 8 '13 at 1:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.