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<p>The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where&nbsp;$x,y&gt;0$. It is given that&nbsp;$y=2$ when&nbsp;$x=1$.</p>
<ol class="abc-list"><li data-list-item-id="p2-tzc-q12-item01" data-list-item-path="p2-tzc-q12-item01"><div class="list-item-paragraph">Use Euler’s method with step size $0.01$ to find an approximation for&nbsp;$y$ when&nbsp;$x=1.05$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item02" data-list-item-path="p2-tzc-q12-item02"><div class="list-item-paragraph">The substitution $\frac1{y^2}=-2z$ is used.</div><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}z}{\mathrm{d}x}-\frac{2z}{x}=1$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="p2-tzc-q12-item03" data-list-item-path="p2-tzc-q12-item03"><div class="list-item-paragraph">Hence solve the differential equation, giving $y$ as a function of $x$.<span class="inline-marks">[7]</span></div></li><li data-list-item-id="p2-tzc-q12-item04" data-list-item-path="p2-tzc-q12-item04"><div class="list-item-paragraph">State the domain and range of the solution.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item05" data-list-item-path="p2-tzc-q12-item05"><div class="list-item-paragraph"><p>By considering the shape of the graph of the solution, state whether the approximation in part <strong>(a)</strong> is an overestimate or an underestimate, giving a reason.</p><span class="inline-marks">[1]</span></div></li></ol>

The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where $x,y>0$. It is given that $y=2$ when $x=1$.

Use Euler’s method with step size $0.01$ to find an approximation for $y$ when $x=1.05$.

Hence solve the differential equation, giving $y$ as a function of $x$.

State the domain and range of the solution.

Euler's Method

Concavity of Graphs

Function Notation, Domain and Range

Solving Differential Equations by Substitution

Linear First-Order Differential Equations and Integrating Factors

IB AA HL May 2026 P2 TZC Q12

Functions

Differential Equations

Applications of Differentiation

<p>The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where&nbsp;$x,y&gt;0$. It is given that&nbsp;$y=2$ when&nbsp;$x=1$.</p>
<p>Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.</p>
<p>The estimated change in $y$ is therefore&nbsp;$h\times\text{current gradient}$. Let&nbsp;$y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is</p>
<p>\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]</p>
<p>At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.</p>
<ol class="abc-list"><li data-list-item-id="p2-tzc-q12-item01" data-list-item-path="p2-tzc-q12-item01"><div class="list-item-paragraph">Take $h=0.01$. First calculate the gradient at $(1,2)$. Use this gradient in the Euler formula to show that the first estimate, at $x_1=1.01$, is $y_1=2.06$.</div><div class="list-item-paragraph">Then calculate $y_2$, showing your substitution and giving its value to 5 decimal places. Continue the process to estimate $y$ when&nbsp;$x=1.05$, giving this estimate to 3 significant figures.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item02" data-list-item-path="p2-tzc-q12-item02"><div class="list-item-paragraph">Using the substitution $u=xy$, show that the differential equation can be written as&nbsp;$\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{u^3}{x^2}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item03" data-list-item-path="p2-tzc-q12-item03"><div class="list-item-paragraph">Hence, by separating the variables and using the given initial condition, solve the differential equation, giving $y$ as a function of $x$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="p2-tzc-q12-item04" data-list-item-path="p2-tzc-q12-item04"><div class="list-item-paragraph">State the largest possible domain of this solution containing $x=1$, and its corresponding range.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item05" data-list-item-path="p2-tzc-q12-item05"><div class="list-item-paragraph">By considering the shape of the solution curve for&nbsp;$1\leq x\leq1.05$ and the straight-line steps in Euler's method, state whether the approximation in part (a) is an overestimate or an underestimate. Give a reason.<span class="inline-marks">[1]</span></div></li></ol>

Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.

The estimated change in $y$ is therefore $h\times\text{current gradient}$. Let $y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is

\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]

At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.

Solving Differential Equations by Separation of Variables

Range of a Function

Product Rule

Domain of a Function

Sketching Solution Curves of Differential Equations

IB AA HL May 2026 P2 TZC Q12 [modified]

Techniques of Differentiation

<p>The variables $x$ and $y$ are related by the differential equation</p>
<p style="text-align: center">\[x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for}\hspace{0.5em} x &gt; 0.\]</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Using the substitution $y = ux^2$, for&nbsp;$x &gt; 0$, show that the differential equation can be reduced to</div><div class="list-item-paragraph" style="text-align: center">\[\frac{\mathrm{d}u}{\mathrm{d}x} = u(10 - u).\]<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Hence show that the general solution for the differential equation</div><div class="list-item-paragraph" style="text-align: center">\[x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for}\hspace{0.5em} x &gt; 0\]</div><div class="list-item-paragraph">can be expressed in the form $y = \frac{10x^2}{Qe^{-10x} + 1}$, where&nbsp;$Q$ is an arbitrary constant.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">Given that the graph of $y = \mathrm{f}(x)$ passes through the point with coordinates $\left(\frac{1}{10}, \frac{1}{10(1 - \mathrm{e}^{-1})}\right)$, find the equation of the graph of $y = \mathrm{f}(x)$.<span class="inline-marks">[2]</span></div></li></ol>

The variables $x$ and $y$ are related by the differential equation

\[x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for } x > 0.\]

2024 MI PRU2 P2 Q5

<ol class="abc-list"><li><div class="list-item-paragraph">Given that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-2xy}{x^2 + y^2}$, use the substitution $y = vx$, where&nbsp;$v$ is a function of $x$ to show that\[x \frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{3v + v^3}{1 + v^2}.\]</div></li><li><div class="list-item-content"><span class="list-item-text">Hence, show that the solution for the given differential equation is $3x^2 y + y^3 = c$ where&nbsp;$c$ is an arbitrary constant.</span></div></li></ol>

Hence, show that the solution for the given differential equation is $3x^2 y + y^3 = c$ where $c$ is an arbitrary constant.

Homogeneous equations

<p>It is given that&nbsp;$2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Use the substitution $y = \ln \left( \frac{u}{x} \right)$ to show that the differential equation can be reduced to $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{f}(u)$, where the function $\mathrm{f}(u)$ is to be found.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$y$ has a minimum value at $x = 3$, solve the differential equation $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$, to find the particular solution for&nbsp;$y$ in terms of $x$.</span><span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of this particular solution.</span><span class="inline-marks">[2]</span></div></li></ol>

It is given that $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$.

Use the substitution $y = \ln \left( \frac{u}{x} \right)$ to show that the differential equation can be reduced to $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{f}(u)$, where the function $\mathrm{f}(u)$ is to be found.

Given that $y$ has a minimum value at $x = 3$, solve the differential equation $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$, to find the particular solution for $y$ in terms of $x$.

Sketch the graph of this particular solution.

Derivatives of Natural Logarithmic Functions

Chain Rule

2025 VJC P1 Q11

<p>It is given that&nbsp;$2xy\frac{\mathrm{d}y}{\mathrm{d}x}=x^2-y^2$ where&nbsp;$x&gt;0$, $y&lt;0$. Using the substitution $u=\frac{y}{x}$, show that the differential equation can be transformed to $\frac{2u}{1-3u^2}\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{x}$. Hence find the general solution of $y$ in terms of $x$.<span class="inline-marks">[6]</span></p>

It is given that $2xy\frac{\mathrm{d}y}{\mathrm{d}x}=x^2-y^2$ where $x>0$, $y<0$. Using the substitution $u=\frac{y}{x}$, show that the differential equation can be transformed to $\frac{2u}{1-3u^2}\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{x}$. Hence find the general solution of $y$ in terms of $x$.

2025 DHS P1 Q1 

<p>The curve $C$, where&nbsp;$y \ge 0$, is such that it satisfies the following differential equation</p>
<p style="text-align: center">\[(1+x^2)y \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy^2 = 0,\]</p>
<p>where&nbsp;$y = 1$ when&nbsp;$x = 0$.</p>
<p>By using the substitution $z = y^2$, solve the differential equation and find the value of $y$ when&nbsp;$x = 1$.<span class="inline-marks">[5]</span></p>

The curve $C$, where $y \ge 0$, is such that it satisfies the following differential equation

\[(1+x^2)y \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy^2 = 0,\]

By using the substitution $z = y^2$, solve the differential equation and find the value of $y$ when $x = 1$.

2020 NYJC P1 Q7

<ol class="abc-list"><li><div class="list-item-paragraph">Consider the differential equation</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}\left( \frac{y}{x} \right), \quad x &gt; 0.$</div><div class="list-item-paragraph">Use the substitution $y = vx$ to show that the general solution of this differential equation is</div><div class="list-item-paragraph" style="text-align: center">$\displaystyle\int \frac{\mathrm{d}v}{\mathrm{f}(v) - v} = \ln x + \text{Constant}.$<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Hence, or otherwise, solve the differential equation</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 + 3xy + y^2}{x^2}, \quad x &gt; 0,$</div><div class="list-item-paragraph">given that $y = 1$ when $x = 1$. Give your answer in the form $y = \mathrm{g}(x)$.<span class="inline-marks">[10]</span></div></li></ol>

2017 IB HL P3 Q4

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">It is given that $x \frac{\mathrm{d}y}{\mathrm{d}x} = xy(\ln x + \ln y) - y$. Using the substitution $w = xy$, show that the differential equation can be transformed to $\frac{\mathrm{d}w}{\mathrm{d}x} = \mathrm{f}(w)$, where the function $\mathrm{f}(w)$ is to be found.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, given that $y = \frac{1}{2}\mathrm{e}^3$ when $x = 2$, solve the differential equation $x \frac{\mathrm{d}y}{\mathrm{d}x} = xy(\ln x + \ln y) - y$, to find $y$ in terms of $x$.</span><span class="inline-marks">[5]</span></div></li></ol>

It is given that $x \frac{\mathrm{d}y}{\mathrm{d}x} = xy(\ln x + \ln y) - y$. Using the substitution $w = xy$, show that the differential equation can be transformed to $\frac{\mathrm{d}w}{\mathrm{d}x} = \mathrm{f}(w)$, where the function $\mathrm{f}(w)$ is to be found.

Hence, given that $y = \frac{1}{2}\mathrm{e}^3$ when $x = 2$, solve the differential equation $x \frac{\mathrm{d}y}{\mathrm{d}x} = xy(\ln x + \ln y) - y$, to find $y$ in terms of $x$.

2024 YIJC P1 Q4

<ol class="i-list"><li><div class="list-item-paragraph">Use the substitution $w = x + y$, show that the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3 - x - y}{1 + x + y}$ can be reduced into $\frac{\mathrm{d}w}{\mathrm{d}x} = \frac{4}{1 + w}$. Solve this differential equation and deduce that the general solution can be written as $(x + y)^2 = 2(3x - y) + A$, where $A$ is an arbitrary constant.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the particular solution for the above differential equation, given that the curve passes through $y$-intercept $(0, 5)$. Hence or otherwise, find the coordinates of the other $y$-intercept.</span><span class="inline-marks">[2]</span></div></li></ol>

Use the substitution $w = x + y$, show that the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3 - x - y}{1 + x + y}$ can be reduced into $\frac{\mathrm{d}w}{\mathrm{d}x} = \frac{4}{1 + w}$. Solve this differential equation and deduce that the general solution can be written as $(x + y)^2 = 2(3x - y) + A$, where $A$ is an arbitrary constant.

Find the particular solution for the above differential equation, given that the curve passes through $y$-intercept $(0, 5)$. Hence or otherwise, find the coordinates of the other $y$-intercept.

2024 MI P2 Q3

<ol class="abc-list"><li><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">It is given that&nbsp;$2xy - x^2 \frac{\mathrm{d}y}{\mathrm{d}x} = 2x^4 + y^2$. Using the substitution $wy = x^2$, show that the differential equation can be transformed to $\frac{\mathrm{d}w}{\mathrm{d}x} = 1 + 2w^2$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, solve the differential equation $2xy - x^2 \frac{\mathrm{d}y}{\mathrm{d}x} = 2x^4 + y^2$.</span><span class="inline-marks">[3]</span></div></li></ol></li><li><div class="list-item-content"><span class="list-item-text">It is given that&nbsp;$u$ satisfies $\frac{\mathrm{d}^2u}{\mathrm{d}x^2} = e^{2x}$ and has a stationary value of $\frac{e^2}{4}$ when&nbsp;$x = 1$. Find $u$ in terms of $x$.</span><span class="inline-marks">[3]</span></div></li></ol>

It is given that $u$ satisfies $\frac{\mathrm{d}^2u}{\mathrm{d}x^2} = e^{2x}$ and has a stationary value of $\frac{e^2}{4}$ when $x = 1$. Find $u$ in terms of $x$.

Second-Order Linear Differential Equations

Solving Differential Equations by Direct Integration

2024 EJC P1 Q5

<p class="paragraph-header">D.E by Method of Substitution</p>
<p>Given that $\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}(x,y)$ and use the substitution of $u=\mathrm{f}(x,y)$ .</p>
<ol class="num-list"><li><div class="list-item-paragraph">Differentiation $u=\mathrm{f}(x,y)$ w.r.t $x$.</div></li><li><div class="list-item-paragraph">Substitute the differentiated equation from (1) into $\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}(x,y)$.</div></li><li><div class="list-item-paragraph">Manipulate the equation into a Separable equation and integrate to solve. </div></li></ol>

Some equations $\dfrac{\mathrm{d}y}{\mathrm{d}x}=G(x,y)$ are neither directly integrable nor immediately separable. A suitable substitution $u=F(x,y)$ can reveal a separable equation.

Treat $y$ as a function of $x$. When $F$ contains both variables, use the multivariable chain rule.

\[\frac{\mathrm{d}u}{\mathrm{d}x}=F_x(x,y)+F_y(x,y)\frac{\mathrm{d}y}{\mathrm{d}x}\]

Total derivative

Set $u=F(x,y)$ and compute $\dfrac{\mathrm{d}u}{\mathrm{d}x}$.

Replace $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ using the original differential equation.

Rewrite every remaining occurrence of $x$ and $y$ in terms of $u$ and the chosen independent variable.

Separate and integrate the resulting equation.

Substitute back for $u$, then apply any initial condition.

If $u=F(x)$ depends only on $x$, the $F_y\dfrac{\mathrm{d}y}{\mathrm{d}x}$ term is absent. If $u=F(y)$, then $\dfrac{\mathrm{d}u}{\mathrm{d}x}=F'(y)\dfrac{\mathrm{d}y}{\mathrm{d}x}$.

Substitution Methods

<p>Show that the differential equation $y\left( \frac{\mathrm{d}y}{\mathrm{d}x}+2y \right)=\frac{x}{{{\mathrm{e}}^{4x+{{x}^{2}}}}}$ can be reduced by the substitution $z={{y}^{2}}{{\mathrm{e}}^{4x}}$ to $\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{2x}{{{\mathrm{e}}^{{{x}^{2}}}}}$. Hence, find the general solution in the form ${{y}^{2}}=\mathrm{f}\left( x \right)$. </p><p style="text-align: right;">[4]</p>

Show that the differential equation $y\left( \frac{\mathrm{d}y}{\mathrm{d}x}+2y \right)=\frac{x}{{{\mathrm{e}}^{4x+{{x}^{2}}}}}$ can be reduced by the substitution $z={{y}^{2}}{{\mathrm{e}}^{4x}}$ to $\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{2x}{{{\mathrm{e}}^{{{x}^{2}}}}}$. Hence, find the general solution in the form ${{y}^{2}}=\mathrm{f}\left( x \right)$. 

2025 YIJC P1 Q1

<p>It is given that $\left( 2xy\frac{\mathrm{d}y}{\mathrm{d}x}+{{y}^{2}} \right)\cos x=\frac{1}{x{{y}^{2}}}$, where $0 < x < \frac{\pi }{2}$ and $y\ne 0$.
</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Using the substitution $u=x{{y}^{2}}$, show that the differential equation can be reduced to $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\sec x}{u}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Hence find the general solution to the differential equation.</div><p style="text-align: right;">[3]</p></li></ol>

It is given that $\left( 2xy\frac{\mathrm{d}y}{\mathrm{d}x}+{{y}^{2}} \right)\cos x=\frac{1}{x{{y}^{2}}}$, where $0 < x < \frac{\pi }{2}$ and $y\ne 0$.


2025 HCI P1 Q3

<p>By using the substitution $y=u+3x$, show that the general solution to the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{9x-3y+1}{3x-y+1}$ can be expressed as $(y-3x-1)^{2}=4x+B$.</p>

By using the substitution $y=u+3x$, show that the general solution to the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{9x-3y+1}{3x-y+1}$ can be expressed as $(y-3x-1)^{2}=4x+B$.

<p>Show that the differential equation $4x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}=(2xy-1)^{2}-5$ may be reduced to $x\frac{\mathrm{d}u}{\mathrm{d}x}=u^{2}-1$ by means of the substitution $u=xy$. Hence, show that the general solution is $y=\frac{1+Cx^{2}}{x-Cx^{3}}$.</p>

Show that the differential equation $4x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}=(2xy-1)^{2}-5$ may be reduced to $x\frac{\mathrm{d}u}{\mathrm{d}x}=u^{2}-1$ by means of the substitution $u=xy$. Hence, show that the general solution is $y=\frac{1+Cx^{2}}{x-Cx^{3}}$.

<p>By using the substitution $y=ux$, solve the differential equation $x\frac{\mathrm{d}y}{\mathrm{d}x}=y+x$, given that $x&gt;0$ and $y=2$ when $x=1$.</p>

By using the substitution $y=ux$, solve the differential equation $x\frac{\mathrm{d}y}{\mathrm{d}x}=y+x$, given that $x>0$ and $y=2$ when $x=1$.

<p>By using the substitution $u=xe^{y}$, find $y$ in terms of $x$ given that $e^{y}\left(x\frac{\mathrm{d}y}{\mathrm{d}x}+1\right)=2x$, and that $y=0$ when $x=2$.</p>

By using the substitution $u=xe^{y}$, find $y$ in terms of $x$ given that $e^{y}\left(x\frac{\mathrm{d}y}{\mathrm{d}x}+1\right)=2x$, and that $y=0$ when $x=2$.

<ol class="i-list"><li><div class="list-item-paragraph">Solve $\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{y}{x}=1$, using the substitution $y=vx$.</div></li><li><div class="list-item-paragraph">Solve $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y-x+1}{y-x+5}$, using the substitution $z=y-x$.</div></li><li><div class="list-item-paragraph">Solve $2x^{3}\frac{\mathrm{d}y}{\mathrm{d}x}=x^{2}y+1$, using the substitution $y=u\sqrt{x}$.</div></li><li><div class="list-item-paragraph">Solve $\frac{\mathrm{d}y}{\mathrm{d}x}+\sin 2y=\sin^{2}y$, using the substitution $z=\tan y$.</div></li></ol>

<p>Find the general solution for the differential equation $(x^{2}-2xy)\frac{\mathrm{d}y}{\mathrm{d}x}=y^{2}-2xy$ by using the substitution $y=ux$.</p>

Find the general solution for the differential equation $(x^{2}-2xy)\frac{\mathrm{d}y}{\mathrm{d}x}=y^{2}-2xy$ by using the substitution $y=ux$.

Math Question Bank