"Differential Equations" — Practice

Question 1

State the order and degree of $\left(\dfrac{d^{3}y}{dx^{3}}ight)^{2}+\left(\dfrac{dy}{dx}ight)^{4}-y=0$.

Accepted Answer

The highest derivative present is the third derivative, so the **order is 3**.

The equation is already polynomial in its derivatives, and that third derivative appears squared, so the **degree is 2**.

The fourth power on the first derivative is a distraction: degree is read off the highest-*order* derivative only, whatever powers the lower ones carry.

Question 2

State the order and degree of $\left(1+\left(\dfrac{dy}{dx}ight)^{2}ight)^{3/2}=k\,\dfrac{d^{2}y}{dx^{2}}$.

Accepted Answer

The highest derivative is the second, so the **order is 2**.

The equation is not yet polynomial in its derivatives because of the power $3/2$. Square both sides to clear it:

$$\left(1+\left(\frac{dy}{dx}ight)^{2}ight)^{3}=k^{2}\left(\frac{d^{2}y}{dx^{2}}ight)^{2}$$

Now it is polynomial, and the second derivative appears squared, so the **degree is 2**.

Reading the degree before clearing the fractional power is the error here.

Question 3

Solve $\dfrac{dy}{dx}=e^{x+y}$ with $y(0)=0$.

Accepted Answer

Write $e^{x+y}=e^{x}e^{y}$, which shows the equation is separable:

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

Integrating both sides: $-e^{-y}=e^{x}+C$.

Applying $y=0$ at $x=0$ gives $-1=1+C$, so $C=-2$.

So $-e^{-y}=e^{x}-2$, that is $e^{-y}=2-e^{x}$.

$$y=-\log\left(2-e^{x}ight)$$

Note this only defines $y$ while $e^{x}<2$, which the solution curve respects.

Question 4

Solve $\dfrac{dy}{dx}=\dfrac{1+y^{2}}{1+x^{2}}$ with $y(0)=1$.

Accepted Answer

Separating gives

$\dfrac{dy}{1+y^{2}}=\dfrac{dx}{1+x^{2}}$

Integrating both sides: $	an^{-1}y=	an^{-1}x+C$.

Applying $y=1$ at $x=0$: $	an^{-1}1=0+C$, so $C=\dfrac{\pi}{4}$.

Hence $	an^{-1}y-	an^{-1}x=\dfrac{\pi}{4}$. Taking the tangent of both sides and using the subtraction formula:

$\dfrac{y-x}{1+xy}=1$, so $y-x=1+xy$.

$$y=\frac{1+x}{1-x}$$

Question 5

Solve $\sec^{2}x\,	an y\,dx+\sec^{2}y\,	an x\,dy=0$.

Accepted Answer

Divide throughout by $	an x\,	an y$:

$\dfrac{\sec^{2}x}{	an x}\,dx+\dfrac{\sec^{2}y}{	an y}\,dy=0$

Each term is now of the form $\dfrac{f'}{f}$, since the derivative of $	an$ is $\sec^{2}$.

Integrating: $\log\lvert	an xvert+\log\lvert	an yvert=\log C$.

Combining the logarithms:

$$	an x\,	an y=C$$

Question 6

Solve $\dfrac{dy}{dx}=\dfrac{y}{x}+	an\dfrac{y}{x}$.

Accepted Answer

The right side depends only on $y/x$, so the equation is homogeneous. Substitute $y=vx$, giving $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$:

$v+x\dfrac{dv}{dx}=v+	an v$

The $v$ cancels, leaving $x\dfrac{dv}{dx}=	an v$, which separates:

$\cot v\,dv=\dfrac{dx}{x}$

Integrating: $\log\lvert\sin vvert=\log\lvert xvert+\log C$, so $\sin v=Cx$.

Substituting back:

$$\sin\!\left(\frac{y}{x}ight)=Cx$$

Question 7

Solve $x\,dy-y\,dx=\sqrt{x^{2}+y^{2}}\,dx$.

Accepted Answer

Rearranging gives $\dfrac{dy}{dx}=\dfrac{y+\sqrt{x^{2}+y^{2}}}{x}$, which is homogeneous.

With $y=vx$, the square root becomes $x\sqrt{1+v^{2}}$, so

$v+x\dfrac{dv}{dx}=v+\sqrt{1+v^{2}}$

The $v$ cancels and it separates:

$\dfrac{dv}{\sqrt{1+v^{2}}}=\dfrac{dx}{x}$

The left side is a standard integral: $\log\left\lvert v+\sqrt{1+v^{2}}ightvert=\log\lvert xvert+\log C$.

So $v+\sqrt{1+v^{2}}=Cx$. Substituting $v=\dfrac{y}{x}$ and multiplying by $x$:

$$y+\sqrt{x^{2}+y^{2}}=Cx^{2}$$

Question 8

Solve $\left(x^{2}+xyight)dy=\left(x^{2}+y^{2}ight)dx$.

Accepted Answer

Every term is of degree 2, so substitute $y=vx$:

$\left(x^{2}+vx^{2}ight)\left(v+x\dfrac{dv}{dx}ight)=x^{2}+v^{2}x^{2}$

Dividing by $x^{2}$: $(1+v)\left(v+x\dfrac{dv}{dx}ight)=1+v^{2}$.

Expanding gives $v+v^{2}+(1+v)x\dfrac{dv}{dx}=1+v^{2}$, so $(1+v)x\dfrac{dv}{dx}=1-v$, which separates:

$\dfrac{(1+v)\,dv}{1-v}=\dfrac{dx}{x}$

Split the left side as $-1+\dfrac{2}{1-v}$ before integrating, giving $-v-2\log\lvert 1-vvert=\log\lvert xvert+c$.

Substituting $v=\dfrac{y}{x}$ and tidying the logarithms:

$$(x-y)^{2}=K\,x\,e^{-y/x}$$

Question 9

Solve $\dfrac{dy}{dx}+y=e^{x}$.

Accepted Answer

Here $P=1$ and $Q=e^{x}$, so

$	ext{IF}=e^{\int 1\,dx}=e^{x}$

Multiplying through, the left side becomes $\dfrac{d}{dx}\left(y\,e^{x}ight)$, so

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

$$y=\frac{e^{x}}{2}+Ce^{-x}$$

Check: differentiating gives $\dfrac{e^{x}}{2}-Ce^{-x}$, and adding $y$ returns $e^{x}$.

Question 10

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

Accepted Answer

With $P=\cot x$:

$	ext{IF}=e^{\int\cot x\,dx}=e^{\log\lvert\sin xvert}=\sin x$

Multiplying through, the left side is $\dfrac{d}{dx}\left(y\sin xight)$, so

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

using the double-angle identity to make the integral standard.

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

Question 11

Solve $\left(1+x^{2}ight)\dfrac{dy}{dx}+2xy=4x^{2}$.

Accepted Answer

Divide through by $1+x^{2}$ to reach standard form:

$\dfrac{dy}{dx}+\dfrac{2x}{1+x^{2}}\,y=\dfrac{4x^{2}}{1+x^{2}}$

Now $P=\dfrac{2x}{1+x^{2}}$, which is of the form $\dfrac{f'}{f}$, so

$	ext{IF}=e^{\log\left(1+x^{2}ight)}=1+x^{2}$

That is the original coefficient, which is a useful signal to watch for. Multiplying restores the equation to its first form, whose left side is $\dfrac{d}{dx}\left[y\left(1+x^{2}ight)ight]$:

$y\left(1+x^{2}ight)=\displaystyle\int 4x^{2}\,dx$

$$y\left(1+x^{2}ight)=\frac{4x^{3}}{3}+C$$

Question 12

Solve $\dfrac{dy}{dx}=\dfrac{1}{x+y^{2}}$.

Accepted Answer

As written this is neither separable nor homogeneous, and $y^{2}$ prevents it from being linear in $y$.

Invert it, treating $x$ as the function of $y$:

$\dfrac{dx}{dy}=x+y^{2}$, that is $\dfrac{dx}{dy}-x=y^{2}$

This is linear in $x$ with $P(y)=-1$, so $	ext{IF}=e^{-y}$.

$x\,e^{-y}=\displaystyle\int y^{2}e^{-y}\,dy$

Integrating by parts twice gives $\displaystyle\int y^{2}e^{-y}dy=-e^{-y}\left(y^{2}+2y+2ight)+C$.

$$x=-\left(y^{2}+2y+2ight)+Ce^{\,y}$$

Question 13

Find the integrating factor of $x\dfrac{dy}{dx}-y=x^{3}$, and solve it.

Accepted Answer

Divide by $x$ to reach standard form:

$\dfrac{dy}{dx}-\dfrac{1}{x}\,y=x^{2}$

So $P=-\dfrac{1}{x}$, and

$	ext{IF}=e^{-\int\frac{dx}{x}}=e^{-\log x}=\dfrac{1}{x}$

Multiplying, the left side is $\dfrac{d}{dx}\left(\dfrac{y}{x}ight)$, so

$\dfrac{y}{x}=\displaystyle\int x^{2}\cdot\frac{1}{x}\,dx=\int x\,dx=\frac{x^{2}}{2}+C$

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

Reading $P$ as $-1$ from the original form would have given the integrating factor $e^{-x}$ and a wrong answer.

Question 14

Solve $x\,dy-y\,dx=0$ by inspection.

Accepted Answer

Divide throughout by $x^{2}$:

$\dfrac{x\,dy-y\,dx}{x^{2}}=0$

The left side is exactly $d\!\left(\dfrac{y}{x}ight)$, so the equation says that this differential vanishes.

An expression whose differential is zero is constant:

$\dfrac{y}{x}=C$, hence

$$y=Cx$$

Separation gives the same answer, but recognising the grouping takes one line and no integration.

Question 15

Solve $x\,dy-y\,dx=\left(x^{2}+y^{2}ight)dx$.

Accepted Answer

The combination $x\,dy-y\,dx$ on the left, with $x^{2}+y^{2}$ on the right, points to one grouping. Divide both sides by $x^{2}+y^{2}$:

$\dfrac{x\,dy-y\,dx}{x^{2}+y^{2}}=dx$

The left side is precisely $d\!\left(	an^{-1}\dfrac{y}{x}ight)$, so

$d\!\left(	an^{-1}\frac{y}{x}ight)=dx$

Integrating both sides directly:

$$	an^{-1}\!\left(\frac{y}{x}ight)=x+C$$

No substitution and no separation was needed; the whole solution was recognising the grouping.

Question 16

A curve passes through $(0,1)$ and its slope at any point equals the sum of the coordinates of that point. Find the curve.

Accepted Answer

The condition says $\dfrac{dy}{dx}=x+y$, which rearranges to

$\dfrac{dy}{dx}-y=x$

This is linear with $P=-1$, so $	ext{IF}=e^{-x}$.

$y\,e^{-x}=\displaystyle\int x\,e^{-x}\,dx$

By parts, $\displaystyle\int xe^{-x}dx=-xe^{-x}-e^{-x}+C=-e^{-x}(x+1)+C$.

So $y\,e^{-x}=-e^{-x}(x+1)+C$, giving $y=-(x+1)+Ce^{x}$.

Applying $y=1$ at $x=0$: $1=-1+C$, so $C=2$.

$$y=2e^{x}-x-1$$

Practice — "Differential Equations"

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