document.write( "Question 1188305: Find the family of solutions of the following equation using integrating factors
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Algebra.Com's Answer #819367 by Edwin McCravy(20064)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "\"3x%5E2y+%2B+%28y%5E5+-+x%5E3%29%2A%22y%27%22+=+0\"\r\n" );
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document.write( "\"3x%5E2y+%2B+%28y%5E5+-+x%5E3%29%2Aexpr%28dy%2Fdx%29+=+0\"\r\n" );
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document.write( "\"3x%5E2y+%2B+y%5E5%2Aexpr%28dy%2Fdx%29+-+x%5E3%2Aexpr%28dy%2Fdx%29+=+0\"\r\n" );
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document.write( "\"3x%5E2y%2Adx+%2B+y%5E5%2Ady+-+x%5E3%2Ady+=+0\"\r\n" );
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document.write( "The trick here is to get the left side into the expression of either the product expression of \r\n" );
document.write( "the differential of uv, which is:\r\n" );
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document.write( "\"u%2Adv+%2B+v%2Adu\"\r\n" );
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document.write( "or the quotient expression for the differential of \"u%2Fv\", which is:\r\n" );
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document.write( "\"%28v%2Adu-u%2Adv%29%2Fv%5E2\"\r\n" );
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document.write( "We notice by inspection that the third term has x3 and the first term has 3x2dx, which is the \r\n" );
document.write( "differential of x3.  So we will get those two terms together and the other term\r\n" );
document.write( "on the other side:\r\n" );
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document.write( "\"3x%5E2y%2Adx+-+x%5E3%2Ady+=+y%5E5%2Ady\"\r\n" );
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document.write( "The product rule isn't going to work because there's a MINUS sign between the terms, not a \r\n" );
document.write( "PLUS sign.  So, we think of trying to make the left side into the form \r\n" );
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document.write( "\"%28v%2Adu-u%2Adv%29%2Fv%5E2\", the differential form for \"u%2Fv\"\r\n" );
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document.write( "The du can be the 3x2dx (with u as x3), and the v can be the y. So we'll write the y first \r\n" );
document.write( "in the first term:  \r\n" );
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document.write( "\"y%2A3x%5E2%2Adx+-+x%5E3%2Ady+=+y%5E5%2Ady\"\r\n" );
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document.write( "Now on the left we have the numerator of the differential form for \"x%5E3%2Fy\".\r\n" );
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document.write( "Now all we need to make the left side into the quotient differential form of \"x%5E3%2Fy%5E%22%22\" is to \r\n" );
document.write( "divide both sides of the equation through by y2:\r\n" );
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document.write( "\"%28y%2A3x%5E2%2Adx+-+x%5E3%2Ady%29%2Fy%5E2+=+expr%28y%5E5%2Fy%5E2%29%2Ady\"\r\n" );
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document.write( "So the integrating factor used here is \"1%2Fy%5E2\".\r\n" );
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document.write( "\"%28y%2A3x%5E2%2Adx+-+x%5E3%2Ady%29%2Fy%5E2\"\"%22%22=%22%22\"\"y%5E3%2Ady\"\r\n" );
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document.write( "Now we can integrate both sides of the equation:\r\n" );
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document.write( "\"int%28%28y%2A3x%5E2%2Adx+-+x%5E3%2Ady%29%2Fy%5E2%29\"\"%22%22=%22%22\"\"int%28y%5E3%2Ady%29\"\r\n" );
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document.write( "The whole left side integrates all together as the quotient \"x%5E3%2Fy\", and, the right side \r\n" );
document.write( "integrates as \"y%5E4%2F4\" and we must add an arbitrary constant:\r\n" );
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document.write( "\"x%5E3%2Fy\"\"%22%22=%22%22\"\"y%5E4%2F4\"\"%22%22%2B%22%22\"\"c%5B1%5D\"\r\n" );
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document.write( "Multiply through by 4y\r\n" );
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document.write( "\"4x%5E3\"\"%22%22=%22%22\"\"y%5E5\"\"%22%22%2B%22%22\"\"Cy\"\r\n" );
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document.write( "[We just write C because 4 times an arbitrary constant is just another arbitrary\r\n" );
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document.write( "Edwin
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