Homogeneous and Linear Equations, with Mixed Practice — Worked Solutions

Ques­tion 1: A homo­ge­neous equa­tion with y/x and x/y#

The prob­lem#

Solve $\displaystyle \frac{dy}{dx} = \frac{y}{x} + \frac{x}{y}$.

Under­stand­ing the prob­lem#

We want the gen­eral solu­tion: a rela­tion between $x$ and $y$ with one arbi­trary con­stant. The right side con­tains only the ratios $\displaystyle \tfrac{y}{x}$ and $\displaystyle \tfrac{x}{y} = \tfrac{1}{y/x}$.

The idea#

Since the right side depends only on $\displaystyle \tfrac{y}{x}$, the equa­tion is homo­ge­neous. Sub­sti­tute $y = vx$, so $\displaystyle \tfrac{dy}{dx} = v + x\tfrac{dv}{dx}$; the vari­ables $v$ and $x$ will then sep­a­rate.

Step-by-step solu­tion#

Step 1. Put $y = vx$. Then $\displaystyle \tfrac{y}{x} = v$ and $\displaystyle \tfrac{x}{y} = \tfrac1v$.

$$\displaystyle v + x\frac{dv}{dx} = v + \frac{1}{v}.$$

Step 2. Can­cel $v$ from both sides.

$$\displaystyle x\frac{dv}{dx} = \frac{1}{v}.$$

Step 3. Sep­a­rate the vari­ables: all $v$ on one side, all $x$ on the other.

$$\displaystyle v\,dv = \frac{dx}{x}.$$

Step 4. Inte­grate both sides.

$$\displaystyle \frac{v^2}{2} = \log\lvert x \rvert + C.$$

Step 5. Put back $\displaystyle v = \tfrac{y}{x}$ and clear frac­tions (mul­ti­ply by $2x^2$; write the new con­stant $2C$ again as $C$).

$$\displaystyle \frac{y^2}{2x^2} = \log\lvert x \rvert + C \;\Longrightarrow\; y^2 = 2x^2\log\lvert x \rvert + Cx^2.$$

Check­ing the answer#

Dif­fer­en­ti­ate $y^2 = 2x^2\log\lvert x\rvert + Cx^2$: $2yy' = 4x\log\lvert x\rvert + 2x + 2Cx$. So $\displaystyle y' = \tfrac{x(2\log\lvert x\rvert + C) + x}{y}$. From the solu­tion, $\displaystyle 2\log\lvert x\rvert + C = \tfrac{y^2}{x^2}$, so $\displaystyle y' = \tfrac{y^2/x + x}{y} = \tfrac{y}{x} + \tfrac{x}{y}$ ✓.

Answer#

$y^2 = 2x^2\log\lvert x \rvert + Cx^2$.

Ques­tion 2: Homo­ge­neous equa­tion in dif­fer­en­tial form#

The prob­lem#

Solve $(x - y)\,dy - (x + y)\,dx = 0$.

Under­stand­ing the prob­lem#

The equa­tion is writ­ten with dif­fer­en­tials. First rewrite it as $\displaystyle \tfrac{dy}{dx} = \ldots$ to see its type.

The idea#

Rear­rang­ing gives $\displaystyle \tfrac{dy}{dx} = \tfrac{x + y}{x - y}$: numer­a­tor and denom­i­na­tor are both of degree $1$, so the equa­tion is homo­ge­neous. Use $y = vx$.

Step-by-step solu­tion#

Step 1. Rearrange.

$$\displaystyle (x - y)\,dy = (x + y)\,dx \;\Longrightarrow\; \frac{dy}{dx} = \frac{x + y}{x - y}.$$

Step 2. Put $y = vx$ and divide the top and bot­tom of the right side by $x$.

$$\displaystyle v + x\frac{dv}{dx} = \frac{1 + v}{1 - v}.$$

Step 3. Iso­late $\displaystyle x\tfrac{dv}{dx}$.

$$\displaystyle x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}.$$

Step 4. Sep­a­rate the vari­ables.

$$\displaystyle \frac{1 - v}{1 + v^2}\,dv = \frac{dx}{x}.$$

Step 5. Split the left side into two stan­dard inte­grals.

$$\displaystyle \int\frac{dv}{1 + v^2} - \int\frac{v\,dv}{1 + v^2} = \int\frac{dx}{x}.$$

The first is $\tan^{-1}v$. For the sec­ond, the numer­a­tor $v$ is half the deriv­a­tive of $1 + v^2$, so it equals $\displaystyle \tfrac12\log(1 + v^2)$.

$$\displaystyle \tan^{-1}v - \tfrac12\log(1 + v^2) = \log\lvert x \rvert + C.$$

Step 6. Put $\displaystyle v = \tfrac{y}{x}$ and com­bine the log­a­rithms: $\displaystyle \tfrac12\log\left(1 + \tfrac{y^2}{x^2}\right) + \log\lvert x \rvert = \tfrac12\log\tfrac{x^2 + y^2}{x^2} + \tfrac12\log x^2 = \tfrac12\log(x^2 + y^2)$.

$$\displaystyle \tan^{-1}\frac{y}{x} = \tfrac12\log(x^2 + y^2) + C.$$

Check­ing the answer#

Dif­fer­en­ti­ate implic­itly: $\displaystyle \tfrac{xy' - y}{x^2 + y^2} = \tfrac{x + yy'}{x^2 + y^2}$, so $xy' - y = x + yy'$, $y'(x - y) = x + y$ ✓.

Answer#

$\displaystyle \tan^{-1}\frac{y}{x} = \frac12\log(x^2 + y^2) + C$.

Ques­tion 3: A lin­ear equa­tion with an expo­nen­tial#

The prob­lem#

Solve $\displaystyle \frac{dy}{dx} - y = e^{2x}$.

Under­stand­ing the prob­lem#

The equa­tion has the form $\displaystyle \tfrac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = -1$ and $Q(x) = e^{2x}$. So it is lin­ear in $y$.

The idea#

Mul­ti­ply by the inte­grat­ing fac­tor $\displaystyle \text{IF} = e^{\int P\,dx}$; then $\displaystyle y\cdot\text{IF} = \int Q\cdot\text{IF}\,dx + C$.

Step-by-step solu­tion#

Step 1. Inte­grat­ing fac­tor.

$$\displaystyle \text{IF} = e^{\int(-1)\,dx} = e^{-x}.$$

Step 2. Apply the for­mula.

$$\displaystyle ye^{-x} = \int e^{2x}\cdot e^{-x}\,dx + C = \int e^{x}\,dx + C = e^x + C.$$

Step 3. Mul­ti­ply by $e^x$.

$$y = e^{2x} + Ce^x.$$

Check­ing the answer#

$y' = 2e^{2x} + Ce^x$, so $y' - y = 2e^{2x} + Ce^x - e^{2x} - Ce^x = e^{2x}$ ✓.

Answer#

$y = e^{2x} + Ce^{x}$.

Ques­tion 4: A lin­ear equa­tion with an ini­tial con­di­tion#

The prob­lem#

Solve $\displaystyle \frac{dy}{dx} + \frac{2y}{x} = x$, given $y(1) = 1$.

Under­stand­ing the prob­lem#

Lin­ear form with $\displaystyle P(x) = \tfrac{2}{x}$, $Q(x) = x$. The con­di­tion $y(1) = 1$ fixes the con­stant, so we want a par­tic­u­lar solu­tion.

The idea#

Inte­grat­ing fac­tor $\displaystyle e^{\int P\,dx}$, gen­eral solu­tion, then sub­sti­tute $x = 1$, $y = 1$ to find $C$.

Step-by-step solu­tion#

Step 1. Inte­grat­ing fac­tor (take $x > 0$, since the con­di­tion is at $x = 1$).

$$\displaystyle \text{IF} = e^{\int\frac{2}{x}dx} = e^{2\log x} = e^{\log x^2} = x^2.$$

Step 2. Apply the for­mula.

$$\displaystyle x^2 y = \int x\cdot x^2\,dx + C = \frac{x^4}{4} + C.$$

Step 3. Use $y(1) = 1$.

$$\displaystyle 1 \cdot 1 = \frac14 + C \;\Longrightarrow\; C = \frac34.$$

Step 4. Divide by $x^2$.

$$\displaystyle y = \frac{x^2}{4} + \frac{3}{4x^2}.$$

Check­ing the answer#

At $x = 1$: $\displaystyle \tfrac14 + \tfrac34 = 1$ ✓. $\displaystyle y' = \tfrac{x}{2} - \tfrac{3}{2x^3}$ and $\displaystyle \tfrac{2y}{x} = \tfrac{x}{2} + \tfrac{3}{2x^3}$; their sum is $x$ ✓.

Answer#

$\displaystyle y = \frac{x^2}{4} + \frac{3}{4x^2}$.

Ques­tion 5: A lin­ear equa­tion with cot x#

The prob­lem#

Solve $\displaystyle \frac{dy}{dx} + y\cot x = 2\cos x$.

Under­stand­ing the prob­lem#

Lin­ear with $P(x) = \cot x$ and $Q(x) = 2\cos x$.

The idea#

Inte­grat­ing fac­tor $\displaystyle e^{\int\cot x\,dx}$, then inte­grate $Q \cdot \text{IF}$; the dou­ble-angle iden­tity $2\sin x\cos x = \sin 2x$ makes the inte­gral easy.

Step-by-step solu­tion#

Step 1. $\displaystyle \int\cot x\,dx = \log\lvert\sin x\rvert$, so

$$\text{IF} = e^{\log\sin x} = \sin x.$$

Step 2. Apply the for­mula.

$$\displaystyle y\sin x = \int 2\cos x\sin x\,dx + C = \int\sin 2x\,dx + C.$$

Step 3. Inte­grate: $\displaystyle \int\sin 2x\,dx = -\tfrac{\cos 2x}{2}$.

$$\displaystyle y\sin x = -\frac{\cos 2x}{2} + C.$$

Check­ing the answer#

Dif­fer­en­ti­ate both sides: $y'\sin x + y\cos x = \sin 2x$. Divide by $\sin x$: $y' + y\cot x = 2\cos x$ ✓.

Answer#

$\displaystyle y\sin x = -\frac{\cos 2x}{2} + C$. (Equiv­a­lently $y\sin x = \sin^2 x + C'$, since $\displaystyle -\tfrac{\cos 2x}{2} = \sin^2 x - \tfrac12$.)

Ques­tion 6: A lin­ear equa­tion after divid­ing through#

The prob­lem#

Solve $\displaystyle (1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$.

Under­stand­ing the prob­lem#

The coef­fi­cient of $\displaystyle \tfrac{dy}{dx}$ is not $1$, so we must divide by $1 + x^2$ to reach the stan­dard lin­ear form.

The idea#

Divide, iden­tify $P$ and $Q$, find the IF.

Step-by-step solu­tion#

Step 1. Divide by $1 + x^2$.

$$\displaystyle \frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{1}{(1 + x^2)^2}.$$

So $\displaystyle P = \tfrac{2x}{1 + x^2}$, $\displaystyle Q = \tfrac{1}{(1 + x^2)^2}$.

Step 2. Inte­grat­ing fac­tor. The numer­a­tor $2x$ is the deriv­a­tive of $1 + x^2$:

$$\displaystyle \int\frac{2x}{1 + x^2}dx = \log(1 + x^2) \;\Longrightarrow\; \text{IF} = 1 + x^2.$$

Step 3. Apply the for­mula.

$$\displaystyle y(1 + x^2) = \int\frac{1}{(1 + x^2)^2}(1 + x^2)\,dx + C = \int\frac{dx}{1 + x^2} + C = \tan^{-1}x + C.$$

Check­ing the answer#

Notice the orig­i­nal left side is exactly $\displaystyle \tfrac{d}{dx}\big[(1 + x^2)y\big]$. Dif­fer­en­ti­at­ing $y(1 + x^2) = \tan^{-1}x + C$ gives $\displaystyle (1 + x^2)y' + 2xy = \tfrac{1}{1 + x^2}$ ✓.

Answer#

$y(1 + x^2) = \tan^{-1}x + C$.

Ques­tion 7: A homo­ge­neous equa­tion with tan(y/x)#

The prob­lem#

Solve $\displaystyle \frac{dy}{dx} = \frac{y}{x} + \tan\frac{y}{x}$.

Under­stand­ing the prob­lem#

The right side depends only on $\displaystyle \tfrac{y}{x}$, so this is homo­ge­neous.

The idea#

Sub­sti­tute $y = vx$, sep­a­rate, inte­grate $\cot v$.

Step-by-step solu­tion#

Step 1. With $y = vx$:

$$\displaystyle v + x\frac{dv}{dx} = v + \tan v \;\Longrightarrow\; x\frac{dv}{dx} = \tan v.$$

Step 2. Sep­a­rate: $\displaystyle \frac{dv}{\tan v} = \cot v\,dv$.

$$\displaystyle \cot v\,dv = \frac{dx}{x}.$$

Step 3. Inte­grate.

$$\log\lvert\sin v\rvert = \log\lvert x\rvert + \log\lvert C\rvert.$$

(Writ­ing the con­stant as $\log\lvert C\rvert$ lets us com­bine the logs.)

Step 4. Com­bine and remove the log­a­rithms.

$$\log\lvert\sin v\rvert = \log\lvert Cx\rvert \;\Longrightarrow\; \sin v = Cx.$$

Step 5. Put back $\displaystyle v = \tfrac{y}{x}$.

$$\displaystyle \sin\frac{y}{x} = Cx.$$

Check­ing the answer#

Dif­fer­en­ti­ate $\displaystyle \sin\tfrac{y}{x} = Cx$: $\displaystyle \cos\tfrac{y}{x}\cdot\tfrac{xy' - y}{x^2} = C = \tfrac{\sin(y/x)}{x}$. So $\displaystyle xy' - y = x\tan\tfrac{y}{x}$, i.e. $\displaystyle y' = \tfrac{y}{x} + \tan\tfrac{y}{x}$ ✓.

Answer#

$\displaystyle \sin\frac{y}{x} = Cx$.

Ques­tion 8: A curve from its slope#

The prob­lem#

Find the equa­tion of the curve pass­ing through $(0, 1)$ whose slope at any point $(x, y)$ is $x + y$.

Under­stand­ing the prob­lem#

"Slope at any point is $x + y$" means $\displaystyle \tfrac{dy}{dx} = x + y$. The curve must also pass through $(0, 1)$, which fixes the con­stant.

The idea#

Rewrite as $\displaystyle \tfrac{dy}{dx} - y = x$: lin­ear with $P = -1$, $Q = x$. The inte­gral $\displaystyle \int xe^{-x}dx$ needs inte­gra­tion by parts.

Step-by-step solu­tion#

Step 1. Stan­dard lin­ear form and IF.

$$\displaystyle \frac{dy}{dx} - y = x, \qquad \text{IF} = e^{-x}.$$

Step 2. Apply the for­mula.

$$\displaystyle ye^{-x} = \int xe^{-x}\,dx + C.$$

Step 3. By parts ($u = x$, $dv = e^{-x}dx$, so $du = dx$, $v = -e^{-x}$):

$$\displaystyle \int xe^{-x}\,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x} - e^{-x}.$$

Step 4. So $ye^{-x} = -xe^{-x} - e^{-x} + C$; mul­ti­ply by $e^x$.

$$y = -x - 1 + Ce^x.$$

Step 5. Use the point $(0, 1)$: $1 = 0 - 1 + C$, so $C = 2$.

$$y = 2e^x - x - 1.$$

Check­ing the answer#

$y(0) = 2 - 0 - 1 = 1$ ✓. $y' = 2e^x - 1$, and $x + y = 2e^x - 1$ ✓.

Answer#

$y = 2e^x - x - 1$.

Ques­tion 9: New­ton's law of cool­ing#

The prob­lem#

New­ton's law of cool­ing says $\displaystyle \frac{dT}{dt} = -k(T - 25)$, where $T$ is the tem­per­a­ture in $^\circ$C and $t$ the time in min­utes. A cup of tea at $85^\circ$C cools to $55^\circ$C in $10$ min­utes. When will it be at $40^\circ$C?

Under­stand­ing the prob­lem#

The room is at $25^\circ$C. The rate of cool­ing is pro­por­tional to how much hot­ter the tea is than the room. We know $T(0) = 85$ and $T(10) = 55$; we must find $t$ with $T(t) = 40$. The con­stant $k$ is unknown and must be found from the data.

The idea#

The equa­tion is sep­a­ra­ble in $T - 25$. Solve it to get an expo­nen­tial, use the two data points, then solve for the required time.

Step-by-step solu­tion#

Step 1. Sep­a­rate and inte­grate.

$$\displaystyle \frac{dT}{T - 25} = -k\,dt \;\Longrightarrow\; \log\lvert T - 25\rvert = -kt + c \;\Longrightarrow\; T - 25 = Ae^{-kt}.$$

Step 2. Use $T(0) = 85$: $85 - 25 = A$, so $A = 60$.

$$T - 25 = 60e^{-kt}.$$

Step 3. Use $T(10) = 55$.

$$\displaystyle 30 = 60e^{-10k} \;\Longrightarrow\; e^{-10k} = \frac12.$$

Step 4. Set $T = 40$.

$$\displaystyle 15 = 60e^{-kt} \;\Longrightarrow\; e^{-kt} = \frac14 = \left(\frac12\right)^2 = \left(e^{-10k}\right)^2 = e^{-20k}.$$

Step 5. Com­pare expo­nents: $kt = 20k$, so $t = 20$.

Check­ing the answer#

The tem­per­a­ture excess halves every $10$ min­utes: $60 \to 30$ (at $10$ min) $\to 15$ (at $20$ min). $25 + 15 = 40$ ✓.

Answer#

The tea reaches $40^\circ$C after $20$ min­utes.

Ques­tion 10: Iden­tify the type and solve#

The prob­lem#

Iden­tify the type of the equa­tion $\displaystyle \frac{dy}{dx} = e^{x - y}$ and solve it.

Under­stand­ing the prob­lem#

Before solv­ing, decide which method applies. Use the law of indices $e^{x - y} = e^x\cdot e^{-y}$.

The idea#

The right side is a prod­uct of a func­tion of $x$ and a func­tion of $y$, so the equa­tion is vari­ables sep­a­ra­ble (not homo­ge­neous, not lin­ear in $y$).

Step-by-step solu­tion#

Step 1. Rewrite: $\displaystyle \frac{dy}{dx} = e^x e^{-y}$.

Step 2. Sep­a­rate: mul­ti­ply both sides by $e^{y}\,dx$.

$$e^{y}\,dy = e^{x}\,dx.$$

Step 3. Inte­grate.

$$e^{y} = e^{x} + C.$$

Check­ing the answer#

Dif­fer­en­ti­ate: $e^y y' = e^x$, so $y' = e^{x - y}$ ✓.

Answer#

Vari­ables sep­a­ra­ble; the solu­tion is $e^{y} = e^{x} + C$.