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Suppose a solution of the differential equation $$(xy^3 + x^2y^7)\frac{dy}{dx} = 1$$ satisfies the initial condition $y \left(\frac{1}{4}\right)=1$ . Then the value of $\dfrac{dy}{dx}$ when $y = −1$ is

(A) $4/3$

(B) $−4/3 $

(C) $16/5$

(D) $−16/5$.

This is not a homogeneus equation so I can't solve it. Do I need to know some special procedure to solve this problem? If the main differential equation is solved then I can solve the problem.

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You know that you have a point on the solution at (1/4,1), because this is given in your problem statement. Using that, can you compute the value of $\frac{dy}{dx}$ at $x=1/4$? – Emily Jul 17 '12 at 15:40
There must be a typo. A regular solution of the differential equation cannot have $y$ changing signs: if $y(x_1) < 0$ and $y(x_2) > 0$. But intermediate value theorem there must exist some $x_3$ in between where $y(x_3) = 0$. But if the solution is regular there (that is, $|y'(x_3)| < \infty$), we must reach a contradiction that $0 = 1$. Hence a regular solution cannot change sign. – Willie Wong Jul 17 '12 at 15:48
Just echoing @WillieWong's comment: Something's not right here. – copper.hat Jul 17 '12 at 16:00
@WillieWong's comment is correct; however, I suspect that the exercise is more about algebraic manipulation of the equation, than computing the actual solution. Alternatively, if there is a typo, then perhaps the condition is supposed to be y = 1, in which case my answer below completely contains the solution. – Emily Jul 17 '12 at 16:02
But $y=1$ makes the problem trivial, since the initial condition has $y=1$. – Robert Israel Jul 17 '12 at 22:21
up vote 3 down vote accepted

Since you want to know $x$ at given values of $y$, it's more convenient to switch dependent and independent variables and write the problem as $$ \dfrac{dx}{dy} = xy^3+x^2 y^7, \ x(1) = 1/4 $$ Now this Bernoulli differential equation actually does have a closed-form solution $$ x \left( y \right) = \left( {\rm e}^{(1-y^4)/4}+4-{y}^{4} \right) ^{-1} $$ But even without that, the fact that the right side of the differential equation is an odd function of $y$ shows that all solutions that pass through $y=0$ must be even functions of $y$. Thus we must have $x(-1) = x(1) = 1/4$. And then you can calculate $dy/dx$ at the point $(x=1/4, y=-1)$ from the original differential equation.

In terms of the original differential equation, of course, we can't have $y$ as a function of $x$ with both $y(1/4) = 1$ and $y(1/4) = -1$. What it means is that the two solutions with initial conditions $y(1/4) = 1$ and $y(1/4) = -1$ collide at a singularity on the line $y=0$, but nevertheless we can regard them as two branches of the same integral curve.

enter image description here

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how you get x(y) . can you explain – Argha Jul 18 '12 at 1:35
Change of variables $u = 1/x$ will give you a linear differential equation. – Robert Israel Jul 18 '12 at 1:55
thank you it really helpful – Argha Jul 18 '12 at 7:28

This is, I think, a neat little problem, so I will provide a more in-depth explanation than my comment. It still smells like homework, so I won't solve it for you, just in case.

The problem states that $y(1/4) = 1$, meaning that when $x = 1/4$, $y = 1$. In that case, we can plug in those values to obtain $\frac{dy}{dx}\mid_{x=1/4} = 16/5$. But that doesn't solve our problem, yet.

Next, we want to figure out what happens when $y=-1$. We don't know what the corresponding value of $x$ is, but we can still plug in $y=-1$ to obtain $\left(x(-1)^3+x^2(-1)^7\right)\frac{dy}{dx}\mid_{y=-1} = 1$.

Now, it's multiple choice, which makes our life easier. We want to turn it into a quadratic equation.

First, multiply both sides by $1/\frac{dy}{dx}\mid_{y=-1}$, and then bring over the $x$'s:

$0 = x^2+x+\frac{1}{\frac{dy}{dx}\mid_{y=-1}}$.

Let $c = \frac{1}{\frac{dy}{dx}\mid_{y=-1}}$. Now, let's use the quadratic equation:

$x = -\frac{1}{2}\pm \frac{\sqrt{1-4c}}{2}$.

Plug in your multiple choice values for $c$.

Immediately, you should be able to eliminate two of them. Why?

For the next two, you should arrive at a contradiction with one of them. Can you see what it is, and why?

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Why negative values can not be an answer i understand. But how i arrive at a contradiction i don't understand so please explain – Argha Jul 17 '12 at 16:34
First, please confirm that the criterion y = -1 as you wrote in your problem statement is not a typo. As the comments to your question indicate, it is mathematically impossible to have a continuous solution to this equation that has values of both y=-1 and y=1. – Emily Jul 17 '12 at 16:38

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