Solve the IVP (differential equations)

Question

A couple questions on my homework I'm struggling with are as follows. If anyone could help me with these that would be awesome

$$\frac{dy}{dx} + 3y = 12x+19,\qquad y(0)=-4.$$

$$\frac{dy}{dx}=(4x+9y)/3x\qquad y(1)=2$$

Sorry but here's another that I seem to be getting wrong. $$\frac{dx}{dy}=(x−8)e^{−2y} \qquad y(8)=\ln(8)$$

The solution I get is $y=\ln(2x^2-16x+64)/2$ but apparently that's wrong

Thanks

Answer 1

Both differential equations are a kind that often yields to the integrating factor method.

We want to realize the left hand side as the result of differentiating of a product of the form $f(x)y$ (then we would get $(f(x)y)' = f(x)y' + f'(x)y$, "explaining" the sum of a term with $y'$ on it, and a term with $y$ on it).

But it doesn't look like that. That's "probably" because we "simplified" after doing the derivative. So we "unsimplify it". We want to find a function $\mu(x)$ (called an "integrating factor") such that $$\mu(x)\left(\frac{dy}{dx} + 3y\right) = \mu(x)\frac{dy}{dx} + 3\mu(x)y$$ is of the form $$\frac{d}{dx}(\mu(x)y) = \mu(x)\frac{dy}{dx} + \mu'(x)y.$$

That means that we need $3\mu(x) = \mu'(x)$. This differential equation is separable, so we can solve it: $$\begin{align*} \frac{d}{dx}\mu &= 3\mu\\ \frac{d\mu}{\mu} &= 3\,dx\\ \int\frac{d\mu}{\mu} &= \int 3\,dx\\ \ln|\mu| &= 3x + C\\ |\mu| &= Ae^{3x},\qquad A\gt 0\\ \mu &= Ae^{3x},\qquad A\neq 0. \end{align*}$$ So we can let $\mu(x) = e^{3x}$. Multiplying the entire equation by $e^{3x}$, we get: $$ \begin{align*} e^{3x}\frac{dy}{dx} + 3e^{3x}y &= e^{3x}(12x+9)\\ \frac{d}{dx}\left(e^{3x}y\right) &= (12x+9)e^{3x}\\ \int d(e^{3x}y) &= \int (12x+9)e^{3x}\,dx. \end{align*}$$ Now do the integral, solve for $y$, and finally plug in the initial value to determine the value of the constant of integration.

The second problem can be solved similarly. You should rewrite the equation in the form $$\frac{dy}{dx} + q(x)y = f(x);$$ then find an "integrating factor" $\mu(x)$ which, when multiplied through, will give you that the left hand side is exactly the derivative of $\mu(x)y$. (You'll need to solve an auxiliary differential equation $\mu'(x) = q(x)\mu(x)$). Then solve for $y$, and use the initial value.

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The third (new) problem would be a simple separable equation if not for the fact that you are interested in the case $x=8$, and at that point you cannot separate (since you cannot divide by $x-8$). Instead, you must again treat it as an integrating factor problem, $$\frac{dx}{dy} -e^{-2y}x = -8e^{-2y}$$ and look for a integrating factor $\mu(y)$ with $\mu'(y) = -\mu(y)e^{-2y}$. This leads to $$\int\frac{d\mu(y)}{\mu(y)} = -\int e^{-2y}\,dy$$ or $$\ln|\mu(y)| = \frac{1}{2}e^{-2y}+C$$ hence $$\mu(y) = e^{\frac{1}{2}e^{-2y}}.$$ So then you have $$\frac{d}{dy}\left( e^{\frac{1}{2}e^{-2y}}x\right) = -8e^{-2y}$$ hence $$e^{\frac{1}{2}e^{-2y}}x = 4e^{-2y}.$$ I don't see how you get to your expression.

Answer 2

For the equation $$\frac{dy}{dx} + 3y = 12x+19,\qquad y(0)=-4,$$ one finds a particular solution, by trying for something of the shape $y=ax+b$, and one adds the general solution of the homogeneous equation. Quickly we get to $y=Ce^{-3x}+4x+5$. The initial condition then gives $C=-9$.

For the equation $$\frac{dy}{dx}=(4x+9y)/3x\qquad y(1)=2,$$ I suggest making the substitution $y=wx$. We then get $$x\frac{dw}{dx}+w=\frac{4+9w}{3}$$ and then $$x\frac{dw}{dx}=\frac{4+6w}{3}.$$ This is a separable differential equation, which I assume you can handle comfortably.

Answer 3

Hint 1: To solve

$$y' + 3 y = 12x + 19,$$

let us use the fact that for any constant $p$ and for any $y$,

$$\left(y e^{px}\right)' = e^{px} (y' + py).$$

Integrating both sides and rewriting, this means

$$y = e^{-px} \int e^{px} (y' + py) dx.$$

One particular choice of $p$ would work well. Then use integration by parts to solve the remaining integral.

Solution: Taking $p = 3$ we get $y = e^{-3x} \int e^{3x} (y' + 3y) dx = e^{-3x} \int e^{3x} (12x + 19) dx$. Using integration by parts, we get $\int x e^{3x} dx = \frac{x}{3} e^{3x} - \frac{1}{3} \int e^{3x} dx = \frac{3x-1}{9} e^{3x} + C$. So we get $y = e^{-3x} (\frac{12x - 4}{3} e^{3x} + \frac{19}{3} e^{3x} + C) = C e^{-3x} + 4x + 5$. From $y(0) = C + 5 = -4$ it follows that $C = -9$ and $y(x) = -9 e^{-3x} + 4x + 5$.

Hint 2: To solve

$$y' - \frac{3}{x} y = \frac{4}{3},$$

try using the fact that for any constant $p$ and for any $y$,

$$\left(y x^{-p}\right)' = x^{-p} \left(y' - \frac{p}{x}y\right).$$

Integrating both sides and rewriting, this means

$$y = x^p \int x^{-p} \left(y' - \frac{p}{x}y\right) dx.$$

Now what would be a good choice of $p$?

Solution: Taking $p = 3$, we get $y = x^3 \int x^{-3} (y' - \frac{3}{x} y) dx = x^3 \int x^{-3} (\frac{4}{3}) = x^3 (\frac{-2}{3} x^{-2} + C) = \frac{-2x}{3} + Cx^3$. Solving $y(1) = \frac{-2}{3} + C = 2$ for $C$ gives us $C = \frac{8}{3}$ and $y(x) = \frac{2x}{3}(4x^2 - 1)$.

Hint 3: For the equation

$$y' = (x - 8) e^{-2y},$$

we can use your suggested substitution $y = \ln(u)$, for $u > 0$, to obtain

$$\frac{u'}{u} = \frac{x - 8}{u^2},$$

or, equivalently,

$$u u' = x - 8.$$

Let us use that for any $u$,

$$(u^2)' = 2 u u'.$$

Integrating both sides and taking square roots, and using that $u$ has to be positive, we get

$$u = \sqrt{2 \int u u' dx}.$$

Now solve for $u$, and use the solution to find $y$. Then use $y(8) = \ln(8)$ to find the remaining unknown constant.

Solution: For $u$ we get $u = \sqrt{2 \int u u' dx} = \sqrt{2 \int (x - 8) dx} = \sqrt{x^2 - 16x + C}$, so $y = \frac{1}{2} \ln(x^2 - 16x + C)$. Since $y(8) = \frac{1}{2} \ln(C - 64) = \ln(8)$ we get $C = 128$, so $y(x) = \frac{1}{2} \ln(x^2 - 16x + 128)$.