ydydx=ex Solve it by separating variables like this: ydy=exdx ∫ydy=∫exdx y22=ex+c

Name: ydydx=ex Solve it by separating variables like this: ydy=exdx ∫ydy=∫exdx y22=ex+c
Uploaded: 2022-02-23T17:22:57+00:00
Description: ydydx=ex Solve it by separating variables like this: ydy=exdx ∫ydy=∫exdx y22=ex+c

Question

ydydx=ex  Solve it by separating variables like this:  ydy=exdx  ∫ydy=∫exdx  y22=ex+c

levurdondishav4 · Accepted Answer

The basic justification is that integration by substitution works, which in turn is justified by the chain rule and the fundamental theorem of calculus.   More specifically, suppose you have:  dydx=g(x)h(y)  Rewrite as:  1h(y)dydx=g(x)  Add the implicit dependency of y on x to obtain  1h(y(x))dydx=g(x)  Now, integrate both sides with respect to x:  ∫1h(y(x))dydxdx=∫g(x)dx  If we do a variable substitution of y for x on the left-hand side (i.e., use the integration by substitution technique), we replace dydxdx with dy. Thus we have  ∫1h(y)dy=∫g(x)dx,  which is the separation of variables formula.   So if you believe integration by substitution, then separation of variables is valid.

abonirali59 · Accepted Answer

&#039;Separation of variables&#039;&#039; in ODE (which has nothing to do with separation of variables in PDE) is a kind of magic that is easy to perform but difficult to justify.Assume that in the given differential equation the quantities x and y are functions of a hidden variable t (time). Then the equation yy′=ex is equivalent to y(t)y(t)x(t)=ex(t), resp.y(t)y(t)=ex(t)x(t)Integrating this from t=0 to t=T one gets12(y2(T)−y02)=ex(T)−ex0where (x0,y0) is the initial condition and T is arbitrary. This means: At any given time the quantities x and y are related by the equation12(y2−y02)=ex−ex0Looking back, one can see that the relation between x and y obtained in this way is exactly the equation obtained by following the recipe given in the books.

alenahelenash · Accepted Answer

maybe its better to think of it as ydydx=ex. the two functions of x are equal, so their indefinite integrals (with respect to x) are equal (i.e. the way you talked about it at the end). moving the differentials around is more of a convenience.

y \frac{dy}{dx}=e^{x} Solve it by ''separating variables'' like this: ydy=e^{x}dx \int ydy=\int e^{x}dx y^{2}/2=e^{x}+c

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