Donald Johnson
2021-12-21
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Stella Calderon

Answered 2021-12-22
Author has **35** answers

The given differential equation is

$(2xy-3{x}^{2})dx+({x}^{2}+y)dy=0$

Comparing it with$Mdx+Mdy=0$

$M=2xy-3{x}^{2}\text{}N={x}^{2}+y$

$\frac{dM}{dy}=2x\text{}\frac{dN}{dx}=2x$

Here,$\frac{dM}{dy}=\frac{dN}{dx}$

Then, it is exact differential equation, so solution is

$\int Mdx+\int N$ (term not contining x) $dy=c$

$\int (2x-3{x}^{2})dx+\int ydy=c$

$y\cdot \frac{2{x}^{2}}{2}-\frac{3{x}^{3}}{3}+\frac{{y}^{2}}{2}=c$

${x}^{2}y-{x}^{3}+\frac{{y}^{2}}{2}=c$

This is solution of given DE.

Comparing it with

Here,

Then, it is exact differential equation, so solution is

This is solution of given DE.

Charles Benedict

Answered 2021-12-23
Author has **32** answers

Assuming $(2xy-3{x}^{2})dx+({x}^{2}+y)dy=0$

Then$2xy+({x}^{2}+y)\frac{dy}{dx}=3{x}^{2}$ (1)

Notice that$d\frac{{x}^{2}y+\frac{{y}^{2}}{2}}{dx}$ is:

$x}^{2}\frac{dy}{dx}+2xy+2\frac{y}{2}\frac{dy}{dx$

$=2xy+({x}^{2}+y)\frac{dy}{dx}$ which is left hand side of (1)

So integrating (1) gives

${x}^{2}y+\frac{{y}^{2}}{2}={x}^{3}+k$

So${y}^{2}+2{x}^{2}y-2{x}^{3}+c=0$ is the solution

Then

Notice that

So integrating (1) gives

So

RizerMix

Answered 2021-12-29
Author has **438** answers

Let

Let

since

The given Equation

Integrating

or,

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I am aware that the solution to a homogeneous first order differential equation of the form $\frac{dy}{dx}=p(x)y$ can be obtained by simply by rearranging to:

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Then it is simply a question of integrating both sides and the answer is straightforward. Now what would happen if RHS had a constant, how a can find a particular solution to this case: $\frac{dy}{dx}=p(x)y+C$

I know that the general solution would be the sum of the homogeneous equation and the particular solution

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Then it is simply a question of integrating both sides and the answer is straightforward. Now what would happen if RHS had a constant, how a can find a particular solution to this case: $\frac{dy}{dx}=p(x)y+C$

I know that the general solution would be the sum of the homogeneous equation and the particular solution

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Which of the following is a TRUE statement?

A)

B)

C)

D)

E)

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Suppose that a dose of $x}_{0$ (initial) gram of a drug is injected into the bloodstream. Assume that the drug leaves the blood and enters the urine at a rate proportional to the amount of drug present in the blood. In addition, assume that half of the drug dose has entered the urine after 0.75 hour. Find the time at which the amount of drug in the blood stream is 5% of the original drug dose $x}_{0$ (initial).

I have the first order differential equation. It's finding$t$ that is the problem.

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