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<h1 style="text-align: center;">The Bernoulli Differential Equation</h1>
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<p class="center"><i>How to solve this special first order differential equation</i></p>
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<p>A <b>Bernoulli equation</b> has this form:</p>
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<p class="center"><span class="large"><span class="intbl"><em>dy</em><strong>dx</strong></span> + P(x)y = Q(x)y<sup>n</sup></span><br>
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where n is any Real Number but not 0 or 1</p>
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<p class="dotpoint">When n = 0 the equation can be solved as a <a href="differential-equations-first-order-linear.html">First Order Linear Differential Equation</a>.</p>
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<p class="dotpoint">When n = 1 the equation can be solved using <a href="separation-variables.html">Separation of Variables</a>.</p>
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<div class="dotpoint">
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<p>For other values of n we can solve it by substituting</p>
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<p class="center large">u = y<sup>1−n</sup></p>
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<p>and turning it into a linear differential equation (and then solve that).</p>
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</div>
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<p><br></p>
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<div class="example">
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<h3><b>Example 1:</b> Solve</h3>
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<p class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> + x<sup>5</sup> y = x<sup>5</sup> y<sup>7</sup></p>
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<p>It is a Bernoulli equation with P(x)=x<sup>5</sup>, Q(x)=x<sup>5</sup>, and n=7, let's try the substitution:</p>
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<div class="fun">
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<p class="center large">u = y<sup>1−n</sup></p>
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<p class="center large">u = y<sup>-6</sup></p>
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<p>In terms of y that is:</p>
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<p class="center large">y = u<sup>(−<span class="intbl"><em>1</em><strong>6</strong></span>)</sup></p>
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<p>Differentiate y with respect to x:</p>
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<p class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> = <span class="intbl"><em>−1</em><strong>6</strong></span> u<sup>(−<span class="intbl"><em>7</em><strong>6</strong></span>)</sup> <span class="intbl"><em>du</em><strong>dx</strong></span></p>
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<p>Substitute <span class="intbl"><em>dy</em><strong>dx</strong></span> and y into the original equation <span class="intbl"><em>dy</em><strong>dx</strong></span> + x<sup>5</sup> y
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= x<sup>5</sup> y<sup>7</sup></p>
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<p class="center large"><span class="intbl"><em>−1</em><strong>6</strong></span>u<sup>(<span class="intbl"><em>−7</em><strong>6</strong></span>)</sup> <span class="intbl"><em>du</em><strong>dx</strong></span> + x<sup>5</sup>u<sup>(<span class="intbl"><em>−1</em><strong>6</strong></span>)</sup> = x<sup>5</sup>u<sup>(<span class="intbl"><em>−7</em><strong>6</strong></span>)</sup></p>
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<p>Multiply all terms by −6u<sup>(<span class="intbl"><em>7</em><strong>6</strong></span>)</sup></p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> − 6x<sup>5</sup>u = −6x<sup>5</sup></p>
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</div>
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<p>The substitution worked! We now have an equation we can hopefully solve.</p>
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<p>Simplify:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> = 6x<sup>5</sup>u − 6x<sup>5</sup></p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> = (u−1)6x<sup>5</sup></p>
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<p>Using <a href="separation-variables.html">separation of variables</a>:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>u−1</strong></span> = 6x<sup>5</sup> dx</p>
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<p>Integrate both sides:</p>
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<p class="center large"><span class="int">∫</span><span class="intbl"><em>1</em><strong>u−1</strong></span> du = <span class="int">∫</span>6x<sup>5</sup> dx</p>
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<p>Gets us:</p>
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<p class="center large">ln(u−1) = x<sup>6</sup> + C</p>
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<p class="center large">u−1 = e<sup>x<sup>6</sup> + C</sup></p>
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<p class="center large">u = e<sup>(x<sup>6</sup> + c)</sup> + 1</p>
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<p>Substitute back y = u<sup>(<span class="intbl"><em>−1</em><strong>6</strong></span>)</sup></p>
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<p class="center large">y = ( e<sup>(x<sup>6</sup> + c)</sup> + 1 )<sup>(<span class="intbl"><em>−1</em><strong>6</strong></span>)</sup></p>
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<p><b>Solved!</b></p>
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<p>And we get these example curves:</p>
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<p class="center"><img src="images/diff-eq-bernoulli-1.svg" alt="Sample Graph" height="340" width="550"> <span class="larger"><br>
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</span></p>
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</div>
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<p>Let's look again at that substitution we did above. We started with:</p>
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<p class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> + x<sup>5</sup>y = x<sup>5</sup>y<sup>7</sup></p>
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<p>And ended with:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> − 6x<sup>5</sup>u = −6x<sup>5</sup></p>
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<p>In fact, <b>in general</b>, we can go straight from</p>
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<p class="center"><span class="large"><span class="intbl"><em>dy</em><strong>dx</strong></span> + P(x)y = Q(x)y<sup>n</sup></span><br>
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n is not 0 or 1</p>
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<p>to:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> + (1−n)uP(x) = (1−n)Q(x)</p>
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<p>Then solve that and finish by putting back <span class="large">y = u<sup>(<span class="intbl"><em>−1</em><strong>n−1</strong></span>)</sup></span></p>
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<p>Let's do that in the next example.</p>
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<div class="example">
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<h3><b>Example 2:</b> Solve</h3>
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<p class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> − <span class="intbl"><em>y</em><strong>x</strong></span> = y<sup>9</sup></p>
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<p>It is a Bernoulli equation with n = 9, P(x) = <span class="intbl"><em>−1</em><strong>x</strong></span> and Q(x) = 1</p>
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<div class="fun">
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<p>Knowing it is a Bernoulli equation we can jump straight to this:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> + (1−n)uP(x) = (1−n)Q(x)</p>
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<p>Which, after substituting n, P(X) and Q(X) becomes:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> + <span class="intbl"><em>8u</em><strong>x</strong></span> = −8</p>
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</div>
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<p>Now let's try to solve that.</p>
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<p>Unfortunately we cannot separate the variables, but the equation is linear and is
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of the form <span class="larger"><span class="intbl"><em>du</em><strong>dx</strong></span> + R(X)u = S(x)</span> with <span class="larger">R(X) = <span class="intbl"><em>8</em><strong>x</strong></span></span> and <span class="larger">S(X) = −8</span></p>
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<p>Which we can solve with steps 1 to 9:</p>
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<p>Step 1: Let u=vw</p>
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<p>Step 2: Differentiate u = vw</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> = v<span class="intbl"><em>dw</em><strong>dx</strong></span> + w<span class="intbl"><em>dv</em><strong>dx</strong></span></p>
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<p>Step 3: Substitute <span class="larger">u = vw</span> and <span class="larger"><span class="intbl"><em>du</em><strong>dx</strong></span> = v <span class="intbl"><em>dw</em><strong>dx</strong></span> + w <span class="intbl"><em>dv</em><strong>dx</strong></span></span> into <span class="larger"><span class="intbl"><em>du</em><strong>dx</strong></span> + <span class="intbl"><em>8u</em><strong>x</strong></span> = −8</span>:</p>
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<p class="center large">v<span class="intbl"><em>dw</em><strong>dx</strong></span> + w<span class="intbl"><em>dv</em><strong>dx</strong></span> + <span class="intbl"><em>8vw</em><strong>x</strong></span> = −8</p>
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<p>Step 4: Factor the parts involving w.</p>
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<p class="center large">v<span class="intbl"><em>dw</em><strong>dx</strong></span> + w(<span class="intbl"><em>dv</em><strong>dx</strong></span> + <span class="intbl"><em>8v</em><strong>x</strong></span>) = −8</p>
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<p>Step 5: Set the part inside () equal to zero, and separate the variables.</p>
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<p class="center large"><span class="intbl"><em>dv</em><strong>dx</strong></span> + <span class="intbl"><em>8v</em><strong>x</strong></span> = 0</p>
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<p class="center large"><span class="intbl"><em>dv</em><strong>v</strong></span> = <span class="intbl"><em>−8dx</em><strong>x</strong></span></p>
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<p>Step 6: Solve this separable differential equation to find v.</p>
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<p class="center large"><span class="int">∫</span><span class="intbl"><em>dv</em><strong>v</strong></span> = − <span class="int">∫</span><span class="intbl"><em>8dx</em><strong>x</strong></span></p>
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<p class="center large">ln(v) = ln(k) − 8ln(x)</p>
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<p class="center large">v = kx<sup>-8</sup></p>
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<p>Step 7: Substitute v back into the equation obtained at step 4.</p>
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<p class="center large">kx<sup>-8</sup> <span class="intbl"><em>dw</em><strong>dx</strong></span> = −8</p>
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<p>Step 8: Solve this to find v</p>
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<p class="center large">kx<sup>-8</sup> dw = −8 dx</p>
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<p class="center large">k dw = −8x<sup>8</sup> dx</p>
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<p class="center large"><span class="int">∫</span> k dw = <span class="int">∫</span> −8x<sup>8</sup> dx</p>
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<p class="center large">kw = <span class="intbl"><em>−8</em><strong>9</strong></span>x<sup>9</sup> + C</p>
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<p class="center large">w = <span class="intbl"><em>1</em><strong>k</strong></span>( <span class="intbl"><em>−8</em><strong>9</strong></span> x<sup>9</sup> + C )</p>
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<p>Step 9: Substitute into u = vw to find the solution to the original equation.</p>
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<p class="center large">u = vw = <span class="intbl"><em>kx<sup>-8</sup></em><strong>k</strong></span>( <span class="intbl"><em>−8</em><strong>9</strong></span> x<sup>9</sup> + C )</p>
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<p class="center large">u = x<sup>-8</sup> ( <span class="intbl"><em>−
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8</em><strong>9</strong></span> x<sup>9</sup> + C )</p>
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<p class="center large">u = <span class="intbl"><em>−8</em><strong>9</strong></span>x + Cx<sup>-8</sup></p>
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<p>Now, the substitution we used was:</p>
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<p class="center large">u = y<sup>1−n</sup> = y<sup>-8</sup></p>
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<p>Which in our case means we need to substitute back y = u<sup>(<span class="intbl"><em>−1</em><strong>8</strong></span>)</sup> :</p>
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<p class="center large">y = ( <span class="intbl"><em>−8</em><strong>9</strong></span> x + c x<sup>-8</sup> ) <sup>(<span class="intbl"><em>−1</em><strong>8</strong></span>)</sup></p>
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<p><b>Done!</b></p>
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<p>And we get this nice family of curves:</p>
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<p class="center"><img src="images/diff-eq-bernoulli-2.svg" alt="Sample Graph" height="340" width="550"></p>
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</div>
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<div class="example">
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<h3><b>Example 3:</b> Solve</h3>
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<p class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> + <span class="intbl"><em>2y</em><strong>x</strong></span> = x<sup>2</sup>y<sup>2</sup>sin(x)</p>
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<p>It is a Bernoulli equation with n = 2, P(x) = <span class="intbl"><em>2</em><strong>x</strong></span> and Q(x) = x<sup>2</sup>sin(x)</p>
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<div class="fun">
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<p>We can jump straight to this:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> + (1−n)uP(x) = (1−n)Q(x)</p>
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<p>Which, after substituting n, P(X) and Q(X) becomes:</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> − <span class="intbl"><em>2u</em><strong>x</strong></span> = − x<sup>2</sup>sin(x)</p>
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</div><br>
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<p>In this case, we cannot separate the variables, but the equation is linear and of the form <span class="larger"> <span class="intbl"><em>du</em><strong>dx</strong></span> + R(X)u = S(x) </span> with <span class="larger">R(X) = <span class="intbl"><em>−2</em><strong>x</strong></span> </span> and <span class="larger">S(X) = −x<sup>2</sup>sin(x)</span></p>
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<p>Solve the steps 1 to 9:</p>
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<p>Step 1: Let u=vw</p>
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<p>Step 2: Differentiate u = vw</p>
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<p class="center large"><span class="intbl"><em>du</em><strong>dx</strong></span> = v<span class="intbl"><em>dw</em><strong>dx</strong></span> + w<span class="intbl"><em>dv</em><strong>dx</strong></span></p>
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<p>Step 3: Substitute <span class="larger">u = vw</span> and <span class="larger"><span class="intbl"><em>du</em><strong>dx</strong></span> = v<span class="intbl"><em>dw</em><strong>dx</strong></span> + w<span class="intbl"><em>dv</em><strong>dx</strong></span></span> into <span class="larger"><span class="intbl"><em>du</em><strong>dx</strong></span> − <span class="intbl"><em>2u</em><strong>x</strong></span> = −x<sup>2</sup>sin(x)</span></p>
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<p class="center large">v<span class="intbl"><em>dw</em><strong>dx</strong></span> + w<span class="intbl"><em>dv</em><strong>dx</strong></span> − <span class="intbl"><em>2vw</em><strong>x</strong></span> = −x<sup>2</sup>sin(x)</p>
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<p>Step 4: Factor the parts involving w.</p>
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<p class="center large">v<span class="intbl"><em>dw</em><strong>dx</strong></span> + w(<span class="intbl"><em>dv</em><strong>dx</strong></span> − <span class="intbl"><em>2v</em><strong>x</strong></span>) = −x<sup>2</sup>sin(x)</p>
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<p>Step 5: Set the part inside () equal to zero, and separate the variables.</p>
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<p class="center large"><span class="intbl"><em>dv</em><strong>dx</strong></span> − <span class="intbl"><em>2v</em><strong>x</strong></span> = 0</p>
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<p class="center large"><span class="intbl"><em>1</em><strong>v</strong></span>dv = <span class="intbl"><em>2</em><strong>x</strong></span>dx</p>
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<p>Step 6: Solve this separable differential equation to find v.</p>
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<p class="center large"><span class="int">∫</span><span class="intbl"><em>1</em><strong>v</strong></span> dv = <span class="int">∫</span><span class="intbl"><em>2</em><strong>x</strong></span> dx</p>
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<p class="center large">ln(v) = 2ln(x) + ln(k)</p>
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<p class="center large">v = kx<sup>2</sup></p>
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<p>Step 7: Substitute u back into the equation obtained at step 4.</p>
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<p class="center large">kx<sup>2</sup><span class="intbl"><em>dw</em><strong>dx</strong></span> = −x<sup>2</sup>sin(x)</p>
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<p>Step 8: Solve this to find v.</p>
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<p class="center large">k dw = −sin(x) dx</p>
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<p class="center large"><span class="int">∫</span>k dw = <span class="int">∫</span>−sin(x) dx</p>
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<p class="center large">kw = cos(x) + C</p>
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<p class="center large">w = <span class="intbl"><em>cos(x) + C</em><strong>k</strong></span></p>
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<p>Step 9: Substitute into u = vw to find the solution to the original equation.</p>
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<p class="center large">u = kx<sup>2</sup><span class="intbl"><em>cos(x) + C</em><strong>k</strong></span></p>
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<p class="center large">u = x<sup>2</sup>(cos(x)+C)</p>
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<p>Finally we substitute back y = u<sup>-1</sup></p>
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<p class="center large">y = <span class="intbl"><em>1</em><strong>x<sup>2</sup> (cos(x)+C)</strong></span></p>
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<p>Which looks like this (example values of C):</p>
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<p class="center"><img src="images/graph-1-x2cosx.svg" alt="1 / (x^2(cos(x)+C))" height="340" width="550"></p>
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</div>
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<p> </p>
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<p class="words">The Bernoulli Equation is attributed to Jacob Bernoulli (1655−1705), one of
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a family of famous Swiss mathematicians.</p>
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<p><br></p>
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<div class="questions">9469, 9470, 9471, 9472, 9473, 9474, 9475, 9476, 9477, 9478</div>
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